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Calculus 1 : Integral Expressions

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Example Questions

Example Question #181 : Integral Expressions

Solve the following integral, where a and b are constants:

$\int abe^{b^2x}dx$

Possible Answers:

$\frac{1}{2}ae^{b^2x}+C$

$ae^{b^2x}+C$

$\frac{ae^{b^2x}}{b}+C$

$\frac{2ae^{b^2x}}{b}+C$

Correct answer:

$\frac{ae^{b^2x}}{b}+C$

Explanation:

Keeping in mind that a and b are only constants, the integral is equal to

$\int abe^{b^2x}dx=\frac{abe^{b^2x}}{b^2}+C=\frac{ae^{b^2x}}{b}+C$

and was found using the following rule:

$\int e^{ax}dx=\frac{e^{ax}}{a}+C$

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Example Question #2262 : Calculus

Evaluate the following integral:

$\int xy\cos(x^2y)dx$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To evaluate the integral, we must make the following substitution:

u=x^2y , du=2xydx

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Rewrite the integral in terms of u and integrate:

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

The integral was found using the following rule:

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

Finally, replace u with our original term:

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

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Example Question #2263 : Calculus

In circuits with a resistor, the equation for voltage drop is given by:

$\frac{d}{dx}V =\frac{d}{dt}Q*R$ , where is voltage, is charge, and is resistance.

Write the equation as an integral expression for .

Possible Answers:

$Q=\int \frac{V}{R} dt$

$Q=\int VR dt$

$Q=\int \frac{\frac{d}{dx}V}{R} dt$

$Q=\int {\frac{d}{dx}V}R dt$

Correct answer:

$Q=\int \frac{\frac{d}{dx}V}{R} dt$

Explanation:

Although this may seem really difficult, we only need to solve for .

$\frac{d}{dx}V =\frac{d}{dt}Q*R$

$\frac{d}{dt}Q=\frac{\frac{\mathrm{d} }{\mathrm{d} x}V}{R}$

To solve for , integrate both sides:

$Q=\int \frac{\frac{d}{dx}V}{R} dt$

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Example Question #182 : Integral Expressions

$\int_{0}^{2}2x^2-5x+1 dx$

Possible Answers:

$-\frac{11}{3}$

$\frac{8}{3}$

$-\frac{5}{3}$

$-\frac{8}{3}$

Correct answer:

$-\frac{8}{3}$

Explanation:

When integrating, remember to add one to the exponent and then put that result on the denominator: $\frac{2x^3}{3}-\frac{5x^2}{2}+x$ . Now evaluate at 2, and then 0. Then subtract the two results. $(\frac{16}{3}-10+2)-0=-\frac{8}{3}$ .

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Example Question #2265 : Calculus

$\int \frac{24x^3-2x^2+4x}{4x}dx$

Possible Answers:

$2x^3-\frac{x^2}{4}+x$

$2x^3+\frac{x^2}{4}+x+C$

$2x^3-\frac{x^2}{4}+4x+C$

$-2x^3-\frac{x^2}{4}+x+C$

$2x^3-\frac{x^2}{4}+x+C$

Correct answer:

$2x^3-\frac{x^2}{4}+x+C$

Explanation:

The first step here is to chop this up into three separate terms and then simplify since we have only one denominator: $\int 6x^2-\frac{1}{2}x+1dx$ . Then, integrate each term, remembering to add one to the exponent and then put that result on the denominator: $6(\frac{x^3}{3})-\frac{1}{2}(\frac{x^2}{2})+x$ . Simplify to get your answer: $2x^3-\frac{x^2}{4}+x+C$ . Remember to add C because it is an indefinite integral.

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Example Question #183 : Integral Expressions

$\int \frac{9+x}{x}dx$

Possible Answers:

$9ln\left | x \right |+x+C$

$9ln\left | x \right |-x+C$

$\frac{1}{9}ln\left | x \right |+x+C$

$ln\left | x \right |+x+C$

Correct answer:

$9ln\left | x \right |+x+C$

Explanation:

First, chop this expression up into two terms: $\int \frac{9}{x}+1dx$ . Then, integrate each term, remembering that when there is a single x on a denominator, the integral is $ln\left | x \right |$ . Therefore, the integration is: $9ln\left | x \right |+x+C$ . Remember to add C because it is an indefinite integral.

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Example Question #183 : Integral Expressions

$\int_{0}^{3}2x^3-3x^2+xdx$

Possible Answers:

108

Correct answer:

Explanation:

First, integrate each term separately. Remember, when integrating, raise the exponent by one and then also put that result on the denominator: $2(\frac{x^4}{4})-3(\frac{x^3}{3})+\frac{x^2}{2}$ . Then evaluate at 3 and then 0. Subtract two results to get: $(\frac{81}{2}-27+\frac{9}{2})-0=18$ .

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Example Question #184 : How To Find Integral Expressions

$\int \sqrt{x}dx$

Possible Answers:

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

$\frac{2}{3}$

$\frac{2}{3}x^{\frac{3}{2}}+C$

$\frac{2}{3}x^2+C$

Correct answer:

$\frac{2}{3}x^{\frac{3}{2}}+C$

Explanation:

To integrate this expression, I would first rewrite it to be $\int x^{\frac{1}{2}}dx$ . Then, add one to the exponent and then put that result on the denominator: $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}$ . Then, simplify to get your answer of $\frac{2}{3}x^{\frac{3}{2}}+C$ . Remember to add C because it is an indefinite integral.

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Calculus 1 : Integral Expressions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #181 : Integral Expressions

Example Question #2262 : Calculus

Example Question #2263 : Calculus

Example Question #182 : Integral Expressions

Example Question #2265 : Calculus

Example Question #183 : Integral Expressions

Example Question #183 : Integral Expressions

Example Question #184 : How To Find Integral Expressions

All Calculus 1 Resources

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