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Calculus 1 : How to find integral expressions

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Example Questions

Example Question #111 : How To Find Integral Expressions

Evaluate the integral:

$\int \frac{dx}{\sqrt{x+5}}$

Possible Answers:

$\ln(x+5)^{\frac{1}{2}}+C$

$\ln(\sqrt{x+5})+C$

$(x+5)^{\frac{1}{2}}+C$

$2\ln(\sqrt{x+5})+C$

Correct answer:

$2\ln(\sqrt{x+5})+C$

Explanation:

To integrate, we must make the following subsitution:

$u=(x+5)^{\frac{1}{2}}, du=\frac{1}{2}(x+5)^{-\frac{1}{2}}dx$

Now, rewrite the integral, and integrate:

$2\int \frac{du}{u}=2\ln \left | u\right |+C$

The following rule was used for the integration:

$\int \frac{dx}{x}=\ln \left | x\right |+C$

Finally, replace u with our original term:

$2\ln(\sqrt{x+5})+C$

Note that the absolute value sign went away, because the square root is always positive.

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Example Question #111 : How To Find Integral Expressions

Evaluate the following integral:

$\int (e^x+x\cos(x^2+3))dx$

Possible Answers:

$e^x+\ln(x^2+3)+C$

$e^x+\sin(x^2+3)+C$

$e^x+\frac{1}{2}\sin(x^2+3)+C$

$e^x+\frac{1}{2}\cos(x^2+3)+C$

Correct answer:

$e^x+\frac{1}{2}\sin(x^2+3)+C$

Explanation:

To integrate, we must split the integral in two:

$\int e^xdx+\int (x\cos(x^2+3))dx$

The first integral is:

$\int e^xdx=e^x+C$

and is identical to the rule used to integrate.

The second integral is performed using the following substitution:

u=x^2+3, du=2xdx

Now, rewrite the integral and integrate:

$\frac{1}{2}\int \cos(u)du=\frac{1}{2}\sin(u)+C$

We used the following rule to integrate:

$\int \cos(x)dx=\sin(x)+C$

Finally, replace u with the original term:

$e^x+\frac{1}{2}\sin(x^2+3)+C$

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Example Question #112 : How To Find Integral Expressions

Evaluate the following integral:

$\int (15\sec(x)+x^2)dx$

Possible Answers:

$\ln\left | \sec(x)+\tan(x)\right |+\frac{x^3}{3}+C$

$15\ln\left | \sec(x)+\tan(x)\right |+\frac{x^3}{3}+C$

$15\tan(x)+\frac{x^3}{3}+C$

$15\ln\left | \sec(x)+\tan(x)\right |+x^3+C$

Correct answer:

$15\ln\left | \sec(x)+\tan(x)\right |+\frac{x^3}{3}+C$

Explanation:

The integral is equal to
$15\ln\left | \sec(x)+\tan(x)\right |+\frac{x^3}{3}+C$

and was found using the following rules:

$\int \sec(x)=\ln\left | \sec(x)+\tan(x)\right |+C$ ,

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ .

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Example Question #111 : How To Find Integral Expressions

Evaluate the following integral:

$\int (x^2+2\cos(x))dx$

Possible Answers:

$\frac{x^3}{3}+2\sin(x)+C$

$x^3+2\sin(x)+C$

$x^3-2\sin(x)+C$

$\frac{x^3}{3}-2\sin(x)+C$

Correct answer:

$\frac{x^3}{3}+2\sin(x)+C$

Explanation:

The integral is equal to

$\frac{x^3}{3}+2\sin(x)+C$

and we used the following rules for integration:

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$ ,

$\int \cos(x)dx=\sin(x)+C$

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Example Question #1161 : Functions

Evaluate the integral:

$\int \left(8x^4+\sec^2(x)+\frac{1}{x^2+1}\right)dx$

Possible Answers:

$8x^5+\tan(x)+\frac{1}{2}\ln(x^2+1)+C$

$\frac{8}{5}x^5+\tan(x)+\arccos(x)+C$

$8x^5+\tan(x)+\arctan(x)+C$

$\frac{8}{5}x^5+\tan(x)+\arctan(x)+C$

Correct answer:

$\frac{8}{5}x^5+\tan(x)+\arctan(x)+C$

Explanation:

The integral was performed using the following rules:

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$ ,

$\int \sec^2(x)dx=\tan(x)+C$ ,

$\int \frac{dx}{x^2+1}=\arctan(x)+C$

Applying the above rules we get the following.

$\frac{8}{5}x^5+\tan(x)+\arctan(x)+C$

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Example Question #113 : How To Find Integral Expressions

Evaluate the following integral:

$\int xe^{3x^2}dx$

Possible Answers:

$\frac{1}{6}e^{3x^2}+C$

$\frac{1}{3}e^{3x}+C$

$e^{3x^2}+C$

$\frac{1}{6}e^{2x^2}+C$

Correct answer:

$\frac{1}{6}e^{3x^2}+C$

Explanation:

To integrate, we must make the following substitution:

u=3x^2, du=6xdx

Now, rewrite the integral and integrate:

$\frac{1}{6}\int e^udu=\frac{1}{6}e^u+C$

The following rule was used to integrate:

$\int e^xdx=e^x+C$

Finally, replace u wiht our original term:

$\frac{1}{6}e^{3x^2}+C$

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Example Question #114 : How To Find Integral Expressions

$\int (e^x+x\cos(x^2))dx$

Possible Answers:

$e^x-\cos(x^2)+C$

$e^x+\frac{1}{2}\sin(x^2)+C$

$e^x+\cos(x^2)+C$

$e^x+\frac{1}{2}\cos(x^2)+C$

Correct answer:

$e^x+\frac{1}{2}\sin(x^2)+C$

Explanation:

The integral can be split into two seperate ones:

$\int e^xdx+\int x\cos(x^2)dx$

The first integral is simple:

$\int e^xdx=e^x+C$

and is identical to the rule used.

The second integral can be performed using the following substitution:

u=x^2, du=2xdx

Rewrite and integrate:

$\frac{1}{2} \int \cos(u)du=\frac{1}{2}\sin(u)+C$

The following rule was used to integrate:

$\int \cos(x)dx=\sin(x)+C$

Finally, replace u with the original term and add the integrals together:

$e^x+\frac{1}{2}\sin(x^2)+C$

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Example Question #118 : How To Find Integral Expressions

Given that the piecewise function f(x)=cos(x) for $x\leq0$ and $f(x)=e^{2x}$ for x>0 , find $\int_{-\frac{\pi }{2}}^{2}f(x)dx$ .

Possible Answers:

$\frac{1-e^2}{2}$

$\frac{1-e^4}{2}$

$\frac{1+e^2}{2}$

$\frac{1+e^4}{4}$

$\frac{1+e^4}{2}$

Correct answer:

$\frac{1+e^4}{2}$

Explanation:

Since we have a piecewise function we will need to set up an integral with two parts. One will f(x)=cos(x) from $-\frac{\pi}{2}$ to zero and the other, $f(x)=e^{2x}$ from zero to two.

Setting up the integral and plugging in the bounds looks like,

$\\ \int_{-\frac{\pi }{2}}^{2}f(x)dx=\int_{-\frac{\pi }{2}}^{0}f(x)dx+\int_{0}^{2}f(x)dx\\ \\=\int_{-\frac{\pi }{2}}^{0}cos(x)dx+\int_{0}^{2}e^{2x}dx\\ \\=sin(x)]\begin{matrix}-\frac{\pi }{2}, \end{matrix}0+\frac{e^{2x}}{2}]\begin{matrix}0, \end{matrix}2\\ \\=[sin(0)-sin(-\frac{\pi }{2})]+[\frac{e^{4}}{2}-\frac{e^0}{2}]\\ \\=[0-(-1)]+[\frac{e^4}{2}-\frac{1}{2}]\\ \\=1-\frac{e^4-1}{2}\\ \\=\frac{2+e^4-1}{2}\\ \\=\frac{1+e^4}{2}$ .

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Example Question #115 : How To Find Integral Expressions

Find the general anti-derivative of ${} x^2 e^x$ .

Possible Answers:

${} \frac{x^3e^x}{3}+C$

${} x^2e^x-2xe^x+2e^x+C$

${} x^2e^x+2xe^x+C$

${} x^2e^x+C$

Correct answer:

${} x^2e^x-2xe^x+2e^x+C$

Explanation:

We have a product of two easy functions, so to get our answer we need to use integration by parts. The general formula for this is:

$\int u dv = uv- \int v du$ .

Choose u=x^2 and dv=e^x dx . Then du=2xdx and v=e^x.

So integration by parts tells us that our antiderivative equals

${} x^2e^x-\int 2xe^x dx.$

Now we must apply integration by parts again, because we still have an integral of a product of two functions. Choose u=2x and dv=e^x dx again, so du=2dx and v=e^x .

Then we have

${} \int 2xe^x dx = 2xe^x-\int2e^x dx.$

But of course, 2e^x is its own antiderivative, because we can pull out constants and e^x is its own anti-derivative.

So combining these two results we get our final answer:

${} \int x^2e^x dx = x^2e^x-2xe^x+2e^x+C.$

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Example Question #115 : How To Find Integral Expressions

Write the integral expression for position p(t) given acceleration a(t) .

Possible Answers:

$p(t)=\int \left[\int a(t)dt\right]dt$

$p(t)=\frac{\mathrm{d^2} }{\mathrm{d} t^2}\int \left[\int a(t)dt\right]dt$

$p(t)=\frac{\mathrm{d^2}a(t) }{\mathrm{d} t^2}$

$p(t)=\frac{\mathrm{d} }\left{\mathrm{d} t}[\int a(t)dt\right]$

Correct answer:

$p(t)=\int \left[\int a(t)dt\right]dt$

Explanation:

The indefinite integral of acceleration is velocity. The indefinite integral of velocity is position. We can write this as:

$v(t)=\int a(t)dt$ , where a(t) is acceleration and v(t) is velocity

$p(t)=\int v(t)dt$ , where p(t) is position

$p(t)=\int \left[\int a(t)dt\right]dt$

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Calculus 1 : How to find integral expressions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #111 : How To Find Integral Expressions

Example Question #111 : How To Find Integral Expressions

Example Question #112 : How To Find Integral Expressions

Example Question #111 : How To Find Integral Expressions

Example Question #1161 : Functions

Example Question #113 : How To Find Integral Expressions

Example Question #114 : How To Find Integral Expressions

Example Question #118 : How To Find Integral Expressions

Example Question #115 : How To Find Integral Expressions

Example Question #115 : How To Find Integral Expressions

All Calculus 1 Resources

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