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## Integral Calculus

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## Integration Tricks

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Consider, for instance, the antiderivative

\[\displaystyle\int e^{- x^2} \, dx.\]

\[I = \int_0^\infty e^{- x^2} \, dx\]

## Odd and Even Functions

\[\int_{-t}^t o(x) \, dx = 0.\]

An even function \(e(x)\) satisfies \(e(x) = e(-x)\) for all \(x\). Therefore, for any \(t\),

\[\int_{-t}^t e(x) \, dx = 2 \int_0^t e(x) \, dx.\]

Evaluate \[\int_{-1}^1 \frac{x^3-2x}{\sqrt{x^4+1}} \, dx.\] Notice that the integrand is an odd function. So, \[\begin{align*} \int_{-1}^1 \frac{x^3-2x}{\sqrt{x^4+1}} \, dx &= \int_{-1}^{1} \frac{(-x)^3-2(-x)}{\sqrt{(-x)^4+1}} \, d(-x) \\ &= - \int_{-1}^1 \frac{x^3-2x}{\sqrt{x^4+1}} \, dx \\ &= 0. \end{align*}\] The final equivalence comes from the fact that the integral is equal to the negative of itself. Therefore, it is \(0\). \(_\square\)

\[ \large \int _{ -1 }^{ 1 }{ \frac { \sin { x } }{ 1+{ x }^{ 2 }+{ x }^{ 4 } } \, dx } = \, ? \]

\[\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx.\]

\[\int_a^b f(x) \, dx = \frac{1}{2} \int_a^b f(x) + f(a+b-x) \, dx.\]

Evaluate \[ \int_3^7 \frac{\ln(x+2)}{\ln\big(24+10x-x^2\big)} \, dx.\] We have \[\begin{align*} \int_3^7 \frac{\ln(x+2)}{\ln\big(24+10x-x^2\big)} \, dx &= \int_3^7 \frac{\ln(x+2)}{\ln(x+2) + \ln(12-x)} \, dx \\\\ &= \frac{1}{2} \int_3^7 \frac{\ln(12-x) + \ln(x+2)}{\ln(x+2) + \ln(12-x)} \, dx \\\\ &= 2.\ _\square \end{align*}\]

Evaluate \[ \int_0^\infty \frac{\ln (2x)}{1 + x^2} \, dx.\] We have \[\begin{align} \int_0^\infty \frac{\ln (2x)}{1 + x^2} \, dx &= \frac{1}{2} \int_0^\infty \frac{\ln(2x) + \ln\big(2x^{-1}\big)}{1 + x^2} \, dx \\\\ &= \frac{2\ln 2}{2} \int_0^\infty \frac{1}{1 + x^2} \, dx \\\\ &= \frac{\pi \ln 2}{2}.\ _\square \end{align}\]

Evaluate the integral \(\displaystyle \int_{0}^{\infty} \frac{\ln x}{x^2+2x+4} \, dx.\)

Round your answer to three decimal places.

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\[\int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1}(x) \, dx = bf(b) - af(a).\]

\[\int_a^b f(x) \, dx - \int_{f(b)}^{f(a)} f^{-1}(x) \, dx = (b-a)f(b) - a\big(f(a)-f(b)\big).\]

Let \(f(x)\) be a one-to-one continuous function such that \(f(1)=4\) and \(f(6)=2\), and assume \(\displaystyle \int_1^6 f(x) \, dx = 15\). Calculate \(\displaystyle \int_2^4 f^{-1}(x) \, dx\). The region bounded by \(f\), \(x = 1\), and \(y = 2\) must have area \(5\), implying the integral in question corresponds to the area \(5 + 1 \cdot (4 - 2) = \boxed{7}\). The above formula for decreasing functions provides the same answer. \(_\square\)

\[ \int u(x) v'(x) \, dx = u(x) v(x) - \int v(x) u'(x) \, dx. \]

This can be thought of as a "backwards" application of the product rule .

Evaluate \[\int_1^7 \ln(1 + x) \, dx.\] Let \(u(x) = \ln(1 + x)\) and \(v(x) = x + 1\). Then, \[\begin{align} \int_1^7 \ln(1 + x) \, dx &= \big[(x+1)\ln(x+1)\big]_1^7 - \int_1^7 \frac{x+1}{x+1} \, dx \\ &= 8\ln 8 - 2\ln 2 - 6 \\ &= 22 \ln 2 - 6.\ _\square \end{align}\]

The (adjusted) beta function \(B(m, \, n)\) is defined for nonnegative integers \(m\) and \(n\) as \[B(m, \, n) = \int_0^1 x^{m}(1-x)^{n} \, dx.\] Find a closed form for \(B(m, \, n)\). Note that \(B(0, \, n) = \frac{1}{n+1}\). Now, suppose \(n\) is fixed, and note for \(m > 0\), \[\begin{align} B(m, \, n) &= \int_0^1 x^m (1-x)^n \, dx \\ &= 0 - \frac{m}{n+1} \int_0^1 x^{m-1} \cdot \left(-(1-x)^{n+1}\right) \, dx \\ &= \frac{m}{n+1} B(m-1, \, n+1). \end{align}\] Thus, \(B(m, \, n) = \frac{m}{n+1} \cdot \frac{m-1}{n+2} \cdots \frac{1}{n+m} B(0, n+m) = \frac{m! n!}{(m+n+1)!}.\) \(_\square\) Learn more about the beta function (with correctly off-set indices) here .

\[ \int_0^1 \left(1-x^2\right)^9 x^9 \, dx \]

Let \(I\) denote the value of the integral above. What is the sum of digits of \(I^{-1}?\)

The first identity is \(\sin^2x+\cos^2x=1.\)

We have \[\begin{align} \int(\sin x+\cos x)^2\, dx &=\int\left(\sin^2x+2\sin x\cos x+\cos^2x\right)\, dx\\ &=\int(1+\sin 2x)\, dx\\ &=x-\dfrac{1}{2}\cos 2x+C, \end{align}\] where \(C\) is the constant of integration.

Here is an example of a less obvious application of that identity:

We have \[\begin{align} \int_0^\frac{\pi}{2}\sin^5x\, dx &=\int_0^\frac{\pi}{2}\sin x\left(1-\cos^2x\right)^2\, dx\\ &=-\int_1^0\left(1-u^2\right)^2\, du\\ &=\int_0^1\left(1-u^2\right)^2\, du\\ &=\frac{8}{15}. \end{align}\]

We have \[\begin{align} \int\frac{\cos 2x}{\sin x+\cos x}\, dx &=\int\frac{\cos^2x-\sin^2x}{\sin x+\cos x}\, dx\\ &=\int(\cos x-\sin x)\, dx\\ &=\sin x+\cos x+C, \end{align}\] where \(C\) is the constant of integration.

Finally, the product-to-sum identities help to solve complex integrals.

We have \[\begin{align} \int\sin 2015x \sin 2016x \, dx &=\dfrac{1}{2}\int(\cos x-\cos 4031x)\, dx\\ &=\dfrac{\sin x}{2}-\dfrac{\sin 4031x}{8062}+C, \end{align}\] where \(C\) is the constant of integration.

Find the value of \(\displaystyle\int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}.\) Use the Weierstrass substitution \(\cos x=\frac{1-t^2}{1+t^2}\) and \(dx=\frac{2dt}{1+t^2}:\) \[\int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}\ =\int_0^1\frac{\frac{2dt}{1+t^2}}{2+\frac{1-t^2}{1+t^2}}=\int_0^1\frac{2}{t^2+3}dt.\] This integral can be finished by a \(u\)-substitution and application of the derivative of arctangent. \(_\square\)

We have \[\int\csc \theta\, d\theta=\int\frac{d\theta}{\sin\theta}=\int\frac{\hspace{3mm} \frac{2dt}{1+t^2}\hspace{3mm} }{\frac{2t}{1+t^2}}=\int\frac{dt}{t}=\ln t+C,\] where \(C\) is the constant of integration. All that's needed is to re-substitute \(t=\tan\frac{\theta}{2}\) to obtain the final value of \[\ln\left(\tan\frac{\theta}{2}\right)+C=-\ln(\cot x+\csc x)+C=\ln\frac{\sin\theta}{\cos\theta+1}+C.\]

Find the value of \(\displaystyle\int_0^1\dfrac{\arcsin\frac{2x}{1+x^2}}{1+x^2}\, dx\). Recall that in the Weierstrass substitution, \(\frac{2x}{1+x^2}=\sin\theta\) and \(\frac{2dx}{1+x^2}=d\theta.\) Then the above can be transformed to \[\displaystyle\int_0^\frac{\pi}{2}\frac{\sin^{-1}(\sin\theta)}{2}\, d\theta=\displaystyle\int_0^\frac{\pi}{2}\frac{\theta}{2}\, d\theta.\] This is the integral of a polynomial, which evaluates to \(\frac{\pi^2}{16}.\) \(_\square\)

\[\int_0^1\frac{x^4\left(1-x^2\right)^5}{\left(1+x^2\right)^{10}}\, dx=A\]

Given the above, find \(\frac{1}{A}.\)

Main article: Taylor Series

Evaluate \[ \int_0^1 \ln x \ln(1 - x) \, dx.\] Note \[ \int_0^1 \ln x \ln(1-x) \, dx = - \int_0^1 \sum_{k = 1}^\infty \frac{x^k \ln x}{k} \, dx.\] Since the monotone convergence theorem applies here, this is equal to \[\begin{align} - \sum_{k = 1}^\infty \int_0^1 \frac{x^k \ln x}{k} \, dx &= \sum_{k = 1}^\infty \frac{1}{k (k+1)^2} \\ &= \sum_{k = 1}^\infty \frac{1}{k(k + 1)} - \sum_{k = 1}^\infty \frac{1}{(k + 1)^2} \\ &= 2 - \frac{\pi^2}{6}.\ _\square \end{align}\]

Main article: Differentiation under the integral sign

\[\dfrac{d}{dx} \displaystyle \int_{g(x)}^{h(x)} f(x,t)dt = f\big(x,h(x)\big)\dfrac{d}{dx}h(x) - f\big(x,g(x)\big)\dfrac{d}{dx}g(x) + \displaystyle \int_{g(x)}^{h(x)}\dfrac{\partial}{\partial x}f(x,t)dt. \]

Compute \[ \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-7x}}{x} \, dx. \] Let \( I(a) = \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-ax}}{x}\, dx.\) Upon differentiating under the integral sign, the equation becomes \[\begin{align} \dfrac{\partial I}{\partial a} &= \displaystyle \int_{0}^{\infty} \dfrac{\partial }{\partial a}\dfrac{e^{-5x} - e^{-ax}}{x}\, dx \\ &= \displaystyle \int_{0}^{\infty} \dfrac{0 -(-xe^{-ax})}{x}\, dx \\ &= \displaystyle \int_{0}^{\infty} e^{-ax}\, dx. \end{align}\] Now, integrating with respect to \(x\) yields the following: \[ \dfrac{\partial I}{\partial a} = \left[\dfrac{e^{-ax}}{a}\right]_{\infty}^{0} \implies \dfrac{\partial I}{\partial a} = \dfrac{1}{a} .\] Integrating both sides with respect to \(a,\) \[ I(a) = \displaystyle \int \dfrac{1}{a}\, da + C \Rightarrow I(a) = \ln a + C, \qquad (1) \] where \(C\) is the constant of integration. Notice that \( I(5) = \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-5x}}{x}\, dx = \int_{0}^{\infty} 0\, dx = 0. \) Substituting these values in \((1)\) gives \[ 0 = \ln 5 + C \implies C = -\ln 5 \implies I(a) = \ln\dfrac{a}{5}.\] To obtain the required integral, substitute \(a = 7\): \[ I(7) = \ln\dfrac{7}{5} \implies \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-7x}}{x}\, dx = \ln\dfrac{7}{5}. \ _\square\]

Compute \[ \displaystyle \int_{0}^{\infty} \dfrac{\sin x}{x}\, dx. \] Let \( I(a) = \displaystyle \int_{0}^{\infty} e^{-ax}\dfrac{\sin x}{x}\, dx. \) Then differentiating under the integral sign gives \[ \dfrac{\partial I(a)}{\partial a} = \displaystyle \int_{0}^{\infty} \dfrac{\partial}{\partial a} e^{-ax}\dfrac{\sin x}{x}\, dx= -\displaystyle \int_{0}^{\infty} e^{-ax}\sin x\, dx. \] Integrating with respect to \(x,\) \[ \dfrac{\partial I(a)}{\partial a} = \left[\dfrac{e^{-ax}\left(a\sin x + \cos x\right)}{a^{2}+1}\right]_{0}^{\infty} = -\dfrac{1}{1+a^{2}}. \] Integrating with respect to \(a,\) \[ I(a) = \displaystyle \int -\dfrac{1}{1+a^{2}}da = -\tan^{-1} a + C, \] where \(C\) is the constant of integration. Now, \[ \displaystyle \lim_{a\rightarrow \infty} = \displaystyle \int_{0}^{\infty} \displaystyle \lim_{a\rightarrow \infty} e^{-ax}\dfrac{\sin x}{x}dx = 0 .\] Using the above information, \[ 0 = -\displaystyle \lim_{a \rightarrow \infty} \tan^{-1} a + C \Rightarrow C = \dfrac{\pi}{2} \Rightarrow I(a) = \dfrac{\pi}{2} - \tan^{-1} a. \] To get our integral, we let \(a = 0\) to obtain \[ I(0) = \dfrac{\pi}{2} - \tan^{-1} 0 = \dfrac{\pi}{2}. \] Therefore, \( \displaystyle \int_{0}^{\infty} \dfrac{\sin x}{x}\, dx = \dfrac{\pi}{2} . \ _\square\)

\[ \int_0^1 x^{50} (\ln x)^{150} \, dx \]

If the value of the integral above is equal to

where \(A,B,\) and \(C\) are positive integers, find the value of \(A+B+C\).

Bonus : Generalize \( \displaystyle \int_0^1 x^{m} (\ln x)^{n} \, dx \).

Find the value of the following integral:

\[ \displaystyle\int_{0}^{\infty} \dfrac{1}{x} \left(\tan^{-1} \pi x - \tan^{-1}x\right)\,dx.\]

Suppose \(a\) and \(b\) are real numbers, \(f\) a function, and \(I = \int_a^b f(x) \, dx.\) If \(f(x) = g(x, s) - g(x, r)\) for some constants \(r\) and \( s\), then \[I = \int_a^b \int_r^s \frac{\partial}{\partial t} g(x,t) \, dt \, dx.\]

Supposing Fubini's theorem holds, the order of integration may be swapped or otherwise altered.

Evaluate \[\int_0^{\infty} \frac{e^{\pi x} - e^{x}}{x(e^{\pi x}+1)(e^{x}+1)} \, dx.\] We have \[ \begin{align*} \int_0^{\infty} \frac{e^{\pi x} - e^{x}}{x(e^{\pi x}+1)(e^{x}+1)} \, dx &= \int_0^\infty \frac{1}{x (e^x + 1)} - \frac{1}{x (e^{\pi x} + 1)} \, dx \\ &= \int_0^\infty \int_1^\pi \frac{e^{tx}}{(e^{tx} + 1)^2} \, dt \, dx \\ &= \int_1^\pi \int_0^\infty \frac{e^{tx}}{(e^{tx} + 1)^2} \, dx \, dt \\ &= \int_1^\pi \left[ - \frac{1}{t (e^{tx} + 1)} \right]_0^\infty \, dt \\ &= \int_1^\pi \frac{1}{(1 + 1)t} \, dt \\ &= \frac{\ln \pi}{2}.\ _\square \end{align*} \]

Suppose \(f\) is a bivariate harmonic function, \((a, \, b)\) is a point in the plane, and \(r\) is a positive real number. Then, \[ \int_0^{2\pi} f(a + r\cos\theta, \, b + r\sin\theta) \, d\theta = 2\pi f(a, \, b).\]

Evaluate \[\int_0^{2\pi} e^{\cos x} \cos(\sin x) \, dx.\] Consider the function \(f(x, \, y) = e^x \cos y.\) Note that \[ \left(\frac{\partial}{\partial x}\right)^2 f + \left(\frac{\partial}{\partial y}\right)^2 f = e^x \cos y + e^x (- \cos y) = 0.\] Therefore, \(f\) is a harmonic function, and it follows that \[\int_0^{2\pi} e^{\cos x} \cos(\sin x) \, dx = 2\pi e^0 \cos(0) = 2\pi.\ _\square\]

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## Section 7.1 : Integration by Parts

Evaluate each of the following integrals.

- \( \displaystyle \int{{4x\cos \left( {2 - 3x} \right)\,dx}}\) Solution
- \( \displaystyle \int_{6}^{0}{{\left( {2 + 5x} \right){{\bf{e}}^{\frac{1}{3}x}}\,dx}}\) Solution
- \( \displaystyle \int{{\left( {3t + {t^2}} \right)\sin \left( {2t} \right)\,dt}}\) Solution
- \( \displaystyle \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}}\) Solution
- \( \displaystyle \int{{{{\bf{e}}^{2z}}\cos \left( {\frac{1}{4}z} \right)\,dz}}\) Solution
- \( \displaystyle \int_{0}^{\pi }{{{x^2}\cos \left( {4x} \right)\,dx}}\) Solution
- \( \displaystyle \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}}\) Solution
- \( \displaystyle \int{{{y^6}\cos \left( {3y} \right)\,dy}}\) Solution
- \( \displaystyle \int{{\left( {4{x^3} - 9{x^2} + 7x + 3} \right){{\bf{e}}^{ - x}}\,dx}}\) Solution

## INTEGRATION PRACTICE QUESTIONS WITH SOLUTIONS

Integrate the following with respect to x

∫ (x + 5) 6 dx = (x + 5) ( 6+1) /(6 +1) + c

∫ 1/(2 - 3x) 4 dx = ∫ (2 - 3x) -4 dx

= (2 - 3x) (-4 + 1) / (-4 + 1) ⋅ (-3) + c

∫ √(3x + 2) dx = ∫ (3x + 2) 1/2 dx

= (3x + 2) 3/2 / (3/2) (3) + c

∫ sin 3x dx = (- cos 3x/3) + c

∫ cos (5 - 11x) dx = sin (5 - 11x) / (-11) + c

∫ cosec 2 (5x - 7) dx = -cot (5x - 7) (1/5) + c

∫ e 8 - 7x dx = e 8 - 7x /(-7) + c

∫ 1/(6 - 4x) dx = (log (6 - 4x))/-4 + c

∫ sec 2 x/5 dx = tan (x/5)/(1/5) + c

∫ cosec (5x + 3) cot (5x + 3) dx

∫ cosec (5x + 3) cot (5x + 3) dx = [- cosec (5x + 3)]/5 + c

∫ 30 sec (2 - 15x) tan (2 - 15x) dx

∫ 30 sec (2 - 15x) tan (2 - 15x) dx

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## How to Integrate

Last Updated: March 3, 2023 References Approved

- This should be familiar because the derivative is also a linear operator; the derivative of a sum is the sum of the derivatives.
- Linearity does not apply just for integrals of polynomials. It applies to any integral where the integrand is a sum of two or more terms.

## Definite Integration

## Antiderivatives of Common Functions

## Integrals of Symmetric Functions

## U-Substitution

## Integration by Parts

## Community Q&A

## You Might Also Like

- ↑ https://www.cuemath.com/calculus/power-rule-of-integration/
- ↑ https://web.ma.utexas.edu/users/m408n/CurrentWeb/LM4-9-4.php
- ↑ https://www.sparknotes.com/math/calcab/introductiontointegrals/section1/
- ↑ https://openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus
- ↑ https://www.chegg.com/learn/calculus/calculus/integrals-of-symmetric-functions-in-calculus

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## Solve Integration questions in 5 SECONDS (BEST TRICKS)

## Basic Integration Problems

You can work on whatever task interests you the most.

I can't believe I have to scan my math problem just to get it checked.

## How to solve Integrals

Integrals are the sum of infinite summands, infinitely small.

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Whenever an integral can be written as:

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if we change t=u(x), the integral transforms in:

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INTEGRATION OF RATIONAL FUNCTIONS

A rational function is any function which can be written as the ratio of two polynomial functions.

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Proper: if the degree of the polynomial divisor is greater than the dividend.

Improper: if the dividend polynomial degree is greater than or equal to the divisor.

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Therefore, the integral of an improper rational function can be written:

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To solve the integral of a rational function is decomposed into a sum of simple fractions:

1) The denominator is decomposed into a product of factors as follows:

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and then obtain the following expression:

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3) The coefficients A, B, …, N, are determined by successively x = a, x = b, etc.

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4) Coefficients obtained, we integrate expression.

Case where the denominator polynomial has multiple roots

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INTEGRATION BY TRIGONOMETRIC SUBSTITUTION

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## You can download the App BioProfe READER to practice this theory with self-corrected exercises .

## How to solve Derivatives

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## 5 Common Data Integration Challenges (And How to Solve Them)

Most businesses handle a tremendous amount of data on a daily basis.

## What's Data Integration?

## Data Integration Challenges

## 1. You have disparate data formats and sources

## 2. Your data isn't available where it needs to be

## 3. You have low-quality or outdated data

This can also happen if you go long periods of time without organizing your databases.

## 4. You're using the wrong integration software for your needs

## 5. You have too much data

## How to Create a Top Data Integration Plan

## 1. Clean up your data

- Clean up duplicates. You can use a de-duplicator tool such as Dedupely , for example. Your applications might also have an option to scan and merge duplicates. This is present in some CRMs and contact management tools, like Google Contacts.
- Scan your tools for outdated or invalid data. This includes emails that keep bouncing back on your email marketing tool, phone numbers with invalid formats, contacts with a misspelled name, etc. Get rid of this data - it's not doing you any good!
- Take a long, hard look at the channels through which you collect data and think about how you can optimize this. For example, if you have a form on a landing page full of unnecessary information fields, remove these from the form, and only collect data you actually need. In addition, make sure that you are compliant with data protection policies such as the General Data Protection Regulation (GDPR).

## 2. Introduce clear processes for data management

## 3. Back up your data

Once your data is cleaned up and backed up, you can move on to the actual integration.

## 4. Choose the right software to assist you with data integration

Before choosing an integration solution, you need to answer the following questions:

- What kind of data needs to be integrated?
- Which of your applications need to be integrated and how?
- How should the data flow within the organization? Do you need a one-way or two-way flow of information?
- Do you need constant, real-time sync or trigger-action data pushes?

## 5. Manage and maintain your data

## Don't forget to share this post!

## What Are Plugins?

## Data Synchronization: What It Is and How to Sync Data for Beginners

## How to Sync Google Contacts with Outlook, iCloud, or a CRM

## 3 Ways to Import Contacts From Google to Your Phone

## How to Export Contacts from Outlook to Gmail

## Why Your Corporation Needs Integrations

## Small Business Data Integration: Strategy and Tools

## How to Sync Gmail Contacts Without Import and Export

## Building a Technology Ecosystem: What You Need to Know

## Software Integrations: A Beginner's Guide

The Tools You Need to Run (and Grow) Better

## Definite Integrals

You might like to read Introduction to Integration first!

## Integration

The symbol for "Integral" is a stylish "S" (for "Sum", the idea of summing slices):

And then finish with dx to mean the slices go in the x direction (and approach zero in width).

## Definite Integral

A Definite Integral has start and end values: in other words there is an interval [a, b].

a and b (called limits, bounds or boundaries) are put at the bottom and top of the "S", like this:

## Example: What is 2 ∫ 1 2x dx

We are being asked for the Definite Integral , from 1 to 2, of 2x dx

First we need to find the Indefinite Integral .

Using the Rules of Integration we find that ∫ 2x dx = x 2 + C

Now calculate that at 1, and 2:

And "C" gets cancelled out ... so with Definite Integrals we can ignore C .

Check : with such a simple shape, let's also try calculating the area by geometry:

Yes, it does have an area of 3.

## Example (continued)

The Definite Integral, from 0.5 to 1.0, of cos(x) dx:

The Indefinite Integral is: ∫ cos(x) dx = sin(x) + C

We can ignore C for definite integrals (as we saw above) and we get:

And another example to make an important point:

The Definite Integral, from 0 to 1, of sin(x) dx:

The Indefinite Integral is: ∫ sin(x) dx = −cos(x) + C

Since we are going from 0, can we just calculate the integral at x=1 ??

What? It is negative ? But it looks positive in the graph.

Because we need to subtract the integral at x=0 . We shouldn't assume it is zero.

So let us do it properly, subtracting one from the other:

But we can have negative regions , when the curve is below the axis:

The Definite Integral, from 1 to 3, of cos(x) dx:

So there is more negative than positive with a net result of −0.700....

## Positive Area

In that case we must calculate the areas separately , like in this example:

## Example: What is the total area between y = cos(x) and the x-axis, from x = 1 to x = 3?

So now we have to do the parts separately:

The curve crosses the x-axis at x = π /2 so we have:

That last one comes out negative, but we want it to be positive, so:

Total area = 0.158... + 0.859... = 1.017 ...

This is very different from the answer in the previous example.

A vertical asymptote between a and b affects the definite integral.

## Area above − area below

The integral adds the area above the axis but subtracts the area below, for a "net value":

## Adding Functions

The integral of f+g equals the integral of f plus the integral of g :

## Reversing the interval

Reversing the direction of the interval gives the negative of the original direction.

## Interval of zero length

When the interval starts and ends at the same place, the result is zero:

## Adding intervals

We can also add two adjacent intervals together:

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## 5.7: Integrals Resulting in Inverse Trigonometric Functions

## Learning Objectives

## Integrals that Result in Inverse Trigonometric Functions

## Rule: Integration Formulas Resulting in Inverse Trigonometric Functions

The following integration formulas yield inverse trigonometric functions:

## Proof of the first formula

Let \( y=\sin^{−1}\frac{x}{a}\). Then \( a \sin y=x\). Now using implicit differentiation, we obtain

\[ \dfrac{d}{dx}(a \sin y)=\dfrac{d}{dx}(x) \nonumber \]

\[ a\cos y\dfrac{dy}{dx}=1 \nonumber \]

\[ \dfrac{dy}{dx}=\dfrac{1}{a\cos y}. \nonumber \]

\[ ∫\dfrac{1}{\sqrt{a^2−u^2}}\,du=\sin^{−1}\left(\frac{u}{a}\right)+C. \nonumber \]

## Example \( \PageIndex{1}\): Evaluating a Definite Integral Using Inverse Trigonometric Functions

Evaluate the definite integral

\[ ∫^{1/2}_0\dfrac{dx}{\sqrt{1−x^2}}. \nonumber \]

## Exercise \(\PageIndex{1}\)

Use the formula in the rule on integration formulas resulting in inverse trigonometric functions.

\( \displaystyle ∫\dfrac{dx}{\sqrt{9−x^2}} \quad=\quad \sin^{−1}\left(\dfrac{x}{3}\right)+C \)

## Example \( \PageIndex{2}\): Finding an Antiderivative Involving an Inverse Trigonometric Function using substitution

\[ ∫\dfrac{dx}{\sqrt{4−9x^2}}.\nonumber \]

Substitute \( u=3x\). Then \( du=3\,dx\) and we have

\[ ∫\dfrac{dx}{\sqrt{4−9x^2}}=\dfrac{1}{3}∫\dfrac{du}{\sqrt{4−u^2}}.\nonumber \]

Applying the formula with \( a=2,\) we obtain

## Exercise \(\PageIndex{2}\)

Find the antiderivative of \(\displaystyle ∫\dfrac{dx}{\sqrt{1−16x^2}}.\)

\( \displaystyle ∫\dfrac{dx}{\sqrt{1−16x^2}} = \dfrac{1}{4}\sin^{−1}(4x)+C\)

## Example \( \PageIndex{3}\): Evaluating a Definite Integral

\[ ∫^{\sqrt{3}/2}_0\dfrac{du}{\sqrt{1−u^2}}\nonumber. \nonumber \]

The format of the problem matches the inverse sine formula. Thus,

## Integrals Resulting in Other Inverse Trigonometric Functions

## Example \( \PageIndex{4}\): Finding an Antiderivative Involving the Inverse Tangent Function

Find the antiderivative of \(\displaystyle ∫\dfrac{1}{9+x^2}\,dx.\)

Apply the formula with \( a=3\). Then,

\[ ∫\dfrac{dx}{9+x^2}=\dfrac{1}{3}\tan^{−1}\left(\dfrac{x}{3}\right)+C. \nonumber \]

## Exercise \(\PageIndex{3}\)

Find the antiderivative of \(\displaystyle ∫\dfrac{dx}{16+x^2}\).

Follow the steps in Example \( \PageIndex{4}\).

\(\displaystyle ∫\dfrac{dx}{16+x^2} = \frac{1}{4}\tan^{−1}\left(\dfrac{x}{4}\right)+C \)

## Example \( \PageIndex{5}\): Applying the Integration Formulas WITH SUBSTITUTION

Find an antiderivative of \(\displaystyle ∫\dfrac{1}{1+4x^2}\,dx.\)

## Exercise \(\PageIndex{4}\)

Use substitution to find the antiderivative of \(\displaystyle ∫\dfrac{dx}{25+4x^2}.\)

\(\displaystyle ∫\dfrac{dx}{25+4x^2} = \dfrac{1}{10}\tan^{−1}\left(\dfrac{2x}{5}\right)+C \)

## Example \( \PageIndex{6}\): Evaluating a Definite Integral

Evaluate the definite integral \(\displaystyle ∫^{\sqrt{3}}_{\sqrt{3}/3}\dfrac{dx}{1+x^2}\).

Use the formula for the inverse tangent. We have

## Exercise \(\PageIndex{5}\)

Evaluate the definite integral \(\displaystyle ∫^2_0\dfrac{dx}{4+x^2}\).

Follow the procedures from Example \(\PageIndex{6}\) to solve the problem.

\(\displaystyle ∫^2_0\dfrac{dx}{4+x^2} = \dfrac{π}{8} \)

## Key Concepts

- Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential and Logarithmic Functions lead directly to integration formulas involving inverse trigonometric functions.
- Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem.
- Substitution is often required to put the integrand in the correct form.

## Key Equations

\(\displaystyle ∫\dfrac{du}{\sqrt{a^2−u^2}}=\sin^{−1}\left(\dfrac{u}{a}\right)+C\)

\(\displaystyle ∫\dfrac{du}{a^2+u^2}=\dfrac{1}{a}\tan^{−1}\left(\dfrac{u}{a}\right)+C\)

\(\displaystyle ∫\dfrac{du}{u\sqrt{u^2−a^2}}=\dfrac{1}{a}\sec^{−1}\left(\dfrac{|u|}{a}\right)+C\)

## Contributors and Attributions

## Integration

## Integration Definition

The concept of integration has developed to solve the following types of problems:

- To find the problem function, when its derivatives are given.
- To find the area bounded by the graph of a function under certain constraints.

## Maths Integration

According to Mathematician Bernhard Riemann,

Let us now try to understand what does that mean:

We know that the differentiation of sin x is cos x.

It is mathematically written as:

So, equation (1) can be written as

Where “C” is the arbitrary constant or constant of integration.

Generally, we can write the function as follow:

(d/dx) [F(x)+C] = f(x), where x belongs to the interval I.

Therefore, the symbolic representation of the antiderivative of a function (Integration) is:

## Integrals in Maths

A definite Integral is represented as:

\(\begin{array}{l}\int_{a}^{b} f(x)dx\end{array} \)

Indefinite integrals are defined without upper and lower limits. It is represented as:

Where C is any constant and the function f(x) is called the integrand.

Also, get some more complete definite integral formulas here.

## Integration Examples

Solve some problems based on integration concept and formulas here.

Example 1: Find the integral of the function: \(\begin{array}{l}\int_{0}^{3} x^{2}dx\end{array} \)

\(\begin{array}{l}Given:\ \int_{0}^{3} x^{2}dx\end{array} \)

\(\begin{array}{l} = \left ( \frac{x^{3}}{3} \right )_{0}^{3}\end{array} \)

Example 2: Find the integral of the function: ∫x 2 dx

∫ (x 2 -1)(4+3x)dx = ∫ 4x 2 +3x 3 -3x-4 dx

∫ (x 2 -1)(4+3x)dx = 4(x 3 /3) + 3(x 4 /4)- 3(x 2 /2) – 4x + C

## Video Lesson on Class 12 Important Calculus Questions

## Frequently Asked Questions on Integration

## What is the use of integration?

The integration is used to find the volume, area and the central values of many things.

## What are the real-life applications of integration?

## What is the fundamental theorem of calculus?

## Mention two different types of integrals in Maths.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Visit BYJU’S for all Maths related queries and study materials

## Leave a Comment Cancel reply

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## Integration

## Integrating simple algebraic expressions

Integration is the inverse process to differentiation. Some people call it anti-differentiation .

\(\frac{{dy}}{{dx}} = a{x^n} \to y = \frac{{a{x^{n + 1}}}}{{n + 1}} + c\) provided \(n \ne - 1\)

Find \(\int {({x^4}} + {x^3})\,\,dx\)

\[\int {({x^4}} + {x^3})\,\,dx\]

\[= \frac{{{x^5}}}{5} + \frac{{{x^4}}}{4} + c\]

Find \(\int {(4{x^3}} + 7{x^{ - 2}})\,\,dx\)

\[\int {(4{x^3}} + 7{x^{ - 2}})\,\,dx\]

\[= \frac{{4{x^4}}}{4} + \frac{{7{x^{ - 1}}}}{{ - 1}} + c\]

Find \(\int {{{(x + 2)}^2}}\,\,dx\)

\[\int {{{(x + 2)}^2}}\,\, dx\]

\[= \int {({x^2}}+ 4x + 4)\,\,dx\]

\[= \frac{{{x^3}}}{3} + \frac{{4{x^2}}}{2} + 4x + c\]

\[= \frac{{{x^3}}}{3} + 2{x^2} + 4x + c\]

Find \(\int {\frac{{x + \sqrt x + \sqrt[3]{x}}}{x}}\,\,dx\)

\[\int {\frac{{x + \sqrt x + \sqrt[3]{x}}}{x}}\,\,dx\]

\[= \int {\frac{{x + {x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}}{x}}\,\,dx\]

\[= \int {\frac{x}{x}} + \frac{{{x^{\frac{1}{2}}}}}{x} + \frac{{{x^{\frac{1}{3}}}}}{x}\,\,dx\]

\[= \int {(1 + {x^{ - \frac{1}{2}}}} + {x^{ - \frac{2}{3}}})\,\,dx\]

\[= x + \frac{{{x^{\frac{1}{2}}}}}{{\frac{1}{2}}} + \frac{{{x^{\frac{1}{3}}}}}{{\frac{1}{3}}} + c\]

\[= x + 2\sqrt x + 3\sqrt[3]{x} + c\]

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## How to solve 'CERTIFICATE_VERIFY_FAILED' in azure function python using Azurite

I've configured Azurite HTTPS Setup using mkcert.

I can connect to Azurite from Azure Storage Explorer.

However, I cannot connect to Azurite from Azure Function (Python).

I refer to azure.storage.blob.BlobClient , and the error message is as follows:

The same code works without HTTPS setup.

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Finding definite integrals using area formulas Get 3 of 4 questions to level up! Practice Finding definite integrals using algebraic properties Get 3 of 4 questions to level up! Practice Definite integrals over adjacent intervals Get 3 of 4 questions to level up! Practice Quiz 2 Level up on the above skills and collect up to 560 Mastery points

Integral calculus gives us the tools to answer these questions and many more. Surprisingly, these questions are related to the derivative, and in some sense, the answer to each one is the opposite of the derivative. Start learning 9,700 Mastery points available in course Course summary Integrals Differential equations Applications of integrals

Many challenging integration problems can be solved surprisingly quickly by simply knowing the right technique to apply. While finding the right technique can be a matter of ingenuity, there are a dozen or so techniques that permit a more comprehensive approach to solving definite integrals.

If you'd like to view the solutions on the web go to the problem set web page, click the solution link for any problem and it will take you to the solution to that problem. Note that some sections will have more problems than others and some will have more or less of a variety of problems.

Section 7.1 : Integration by Parts Evaluate each of the following integrals. ∫ 4xcos(2 −3x)dx ∫ 4 x cos ( 2 − 3 x) d x Solution ∫ 0 6 (2 +5x)e1 3xdx ∫ 6 0 ( 2 + 5 x) e 1 3 x d x Solution ∫ (3t+t2)sin(2t)dt ∫ ( 3 t + t 2) sin ( 2 t) d t Solution ∫ 6tan−1( 8 w) dw ∫ 6 tan − 1 ( 8 w) d w Solution ∫ e2zcos(1 4 z)dz ∫ e 2 z cos ( 1 4 z) d z Solution

Integration is a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points and many useful things. But it is easiest to start with finding the area between a function and the x-axis like this: What is the area? Slices

INTEGRATION PRACTICE QUESTIONS WITH SOLUTIONS Question 1 : Integrate the following with respect to x ∫ (x + 5) 6 dx Solution : ∫ (x + 5) 6 dx = (x + 5) (6+1) / (6 +1) + c = (x + 5) 7 /7 + c Question 2 : Integrate the following with respect to x ∫ 1/ (2 - 3x) 4 dx Solution : ∫ 1/ (2 - 3x) 4 dx = ∫ (2 - 3x) -4 dx

The Integral Calculator lets you calculate integrals and antiderivatives of functions online — for free! Our calculator allows you to check your solutions to calculus exercises. It helps you practice by showing you the full working (step by step integration). All common integration techniques and even special functions are supported.

Integration is a linear operator, which means that the integral of a sum is the sum of the integrals, and the coefficient of each term can be factored out, like so: This should be familiar because the derivative is also a linear operator; the derivative of a sum is the sum of the derivatives.

Learn about integrals using our free math solver with step-by-step solutions.

Symbolab is the best integral calculator solving indefinite integrals, definite integrals, improper integrals, double integrals, triple integrals, multiple integrals, antiderivatives, and more. What does to integrate mean? Integration is a way to sum up parts to find the whole.

How to solve integration questions - 1. General Tips 2. Taking Notes 3. Getting Help 4. Doing Homework 5. Problem Solving 6. Studying For an Exam 7. Taking an

Here are some questions based on the integration concept with solutions. 1. Integrate 1/ (1+x2) for limit [0,1]. Solution: I = ∫ 0 1 1 1 + x 2 d x = [ tan − 1 x] 0 1 = [ tan − 1 1 − tan − 1 0] = [ π 4 − 0] = π 4 ∫ 0 1 1 1 + x 2 d x = π 4 2. Find the value of ∫2x cos (x2 - 5). Solution: Let, I = ∫2xcos (x 2 - 5).dx Let x 2 - 5 = t ….. (1) 2x.dx = dt

To solve the integral of a rational function is decomposed into a sum of simple fractions: 1) The denominator is decomposed into a product of factors as follows: 2) Is then written and then obtain the following expression: 3) The coefficients A, B, …, N, are determined by successively x = a, x = b, etc. For example:

1. Clean up your data. Cleaning up your data is an absolutely critical step to take before even thinking about integrating your software ecosystem. The first thing you need to do is to take a look at your existing databases and: Clean up duplicates. You can use a de-duplicator tool such as Dedupely, for example.

Example: What is2∫12x dx. We are being asked for the Definite Integral, from 1 to 2, of 2x dx. First we need to find the Indefinite Integral. Using the Rules of Integration we find that ∫2x dx = x2 + C. Now calculate that at 1, and 2: At x=1: ∫ 2x dx = 12 + C. At x=2: ∫ 2x dx = 22 + C. Subtract:

Exercise 5.7. 1. Find the indefinite integral using an inverse trigonometric function and substitution for ∫ d x 9 − x 2. Hint. Answer. In many integrals that result in inverse trigonometric functions in the antiderivative, we may need to use substitution to see how to use the integration formulas provided above.

Maths Integration. In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale. Calculation of small addition problems is an easy task which we can do manually or by using ...

Solution. This just means, integrate \ ( {x^2}\) with respect to \ (x\). Remember, add one to the power and divide by the new power. The \ (+ c\) appears because when you differentiate a constant ...

It can also provide insights into your decision-making process and help the interviewer determine whether your beliefs and personality align with the employer's organisational culture. Here are six examples of problem-solving interview questions, sample answers and tips for answering them: 1. Tell me about a time you solved a problem without ...

3 answers. Install before the ODBC driver version 17 manually: Download ODBC Driver for SQL Server. If Microsoft ODBC Driver 17 for SQL Server was already installed, but SQL Server 2022 installation said it could not find it. Then try to uninstall Microsoft ODBC Driver 17 for SQL Server, and let ISO install its own.

An Azure tool that is used to manage cloud storage resources on Windows, macOS, and Linux.