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Calculus 1 : How to find integral expressions

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Example Questions

Example Question #21 : Equations

Set up the integral to find the area of a region bounded by the curve y=10-x^2 from x=-3 to x=3 .

Possible Answers:

$A=\int_{-3}^{3} \sqrt{10-y}\: dy$

$A=\int_{0}^{3} 10-x^2\: dx$

$A=\int_{-3}^{3} 10-x^2\: dx$

$A=-\int_{-3}^{3} 10-y^2\: dy$

$A=\int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{10-y}\: dy$

Correct answer:

$A=\int_{-3}^{3} 10-x^2\: dx$

Explanation:

In order to find the area of the region under the curve of the parabola, simply integrate from the given bounds. Ensure that there is no negative area within the bounds by setting the parabola equal to zero.

y=10-x^2

0=10-x^2

x^2=10

$x=\pm \sqrt{10}$

After testing the numbers to the left and right of the critical values, the positive region is in between bounds.

Set up the integral.

$A=\int_{-3}^{3} 10-x^2\: dx$

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Example Question #21 : How To Find Integral Expressions

Evaluate:

$\\ \int x^2+x+1$

Possible Answers:

2x^2+x+C

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

x^3+x^2+x+C

x^3+x^2+x

$\frac{1}{3}x^3+\frac{1}{2}x^2+x$

Correct answer:

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

Explanation:

We get this by applying the inverse power rule to each polynomial term.

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

Therefore we get,

$\\ \int x^2+x+1=\frac{1}{3}x^3+\frac{1}{2}x^2+x+C$

Since this is an indefinite integral, we need to add a constant term at the end of the function.

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Example Question #21 : How To Find Integral Expressions

Evaluate:

$\int_{0}^{1} \sin(x)+x^4 dx$

Possible Answers:

$\cos(1)$

$\frac{6}{5}-\cos(1)$

$\cos(1)-\frac{4}{5}$

$\frac{6}{5}$

Correct answer:

$\frac{6}{5}-\cos(1)$

Explanation:

In order to evaluate the integral, we need to use the inverse power law and the integral of trig functions.

Remember that the inverse power rule is

$\int ax^b dx=\frac{a}{b+1}x^{b+1}$ and the integral of sine is a negative cosine.

$\int_{0}^{1} \sin(x)+x^4 dx$

$\\ \\ =\left[-\cos(x)+\frac{1}{5}x^5\right]_{0}^1 \\ \\ =\left[-\cos(1)+\frac{1}{5}\right]- \left[-\cos(0)+0 \right] \\ \\ =-\cos(1)+\frac{1}{5}+1=\frac{6}{5}-\cos(1)$

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Example Question #1074 : Functions

Evaluate:

$\int x^3+x^2+x+2\ dx$

Possible Answers:

x^4+x^3+x^2+x +C

x^4+x^3+x^2+x

$\frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+2x +C$

$\frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+2x$

Correct answer:

$\frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+2x +C$

Explanation:

In order to evaluate the integral, we need to use the inverse power rule.

Remember that the inverse power rule is

$\int ax^b dx=\frac{a}{b+1}x^{b+1}$

$\int x^3+x^2+x+2\ dx$

$= \frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+2x +C$

Remember to have the constant term since this is an indefinite integral.

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Example Question #21 : How To Find Integral Expressions

Evalulate:

$\int_0^\pi \sin(\theta) d\theta$

Possible Answers:

$\frac{\pi}{2}$

$\pi$

Correct answer:

Explanation:

We evaluate the definite integral by using the definition of the integral for $\sin(\theta)$ .

$\int_0^\pi \sin(\theta) d\theta=-\cos(\theta)\bigg|_{0}^{\pi}$

$-\cos(\theta)\bigg|_{0}^{\pi}=[-\cos(\pi)]-[-\cos(0)]=-(-1)+1=2$

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Example Question #22 : How To Find Integral Expressions

A given derivative f'(x) is defined as $f'(x)=3x^{2}+4x$ . What is f(x) ?

Possible Answers:

f(x)=6x+4

$f(x)=x^{3}+2x^{2}$

None of the above

$f(x)=x^{3}+2x^{2}+C$

f(x)=6x+4+C

Correct answer:

$f(x)=x^{3}+2x^{2}+C$

Explanation:

In order to find f(x) , we must take the indefinite integral of the given derivative $f'(x)=3x^{2}+4x$ .

Noting that the power rule for integrals is

$\int x^{n}=\frac{x^{n+1}}{n+1}+C$ for all $n\neq-1$ and with an arbitrary constant of integration ,

we can determine that

$f(x)=x^{3}+2x^{2}+C$ .

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Example Question #1077 : Functions

A given derivative f'(x) is defined as $f'(x)=6x^{2}+10x$ . What is f(x) ?

Possible Answers:

f'(x)=12x+10

$f(x)=2x^{3}+5x^{2}$

f'(x)=12x+10+C

None of the above

$f(x)=2x^{3}+5x^{2}+C$

Correct answer:

$f(x)=2x^{3}+5x^{2}+C$

Explanation:

In order to find f(x) , we must take the indefinite integral of the given derivative $f'(x)=6x^{2}+10x$ .

Noting that the power rule for integrals is

$\int x^{n}=\frac{x^{n+1}}{n+1}+C$ for all $n\neq-1$ and with an arbitrary constant of integration ,

we can determine that

$f(x)=2x^{3}+5x^{2}+C$ .

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Example Question #1078 : Functions

A given derivative f'(x) is defined as $f'(x)=x^{2}+4x$ . What is f(x) ?

Possible Answers:

f(x)=2x+4+C

f(x)=2x+4

$f(x)=\frac{1}{3}x^{3}+2x^{2}$

None of the above

$f(x)=\frac{1}{3}x^{3}+2x^{2}+C$

Correct answer:

$f(x)=\frac{1}{3}x^{3}+2x^{2}+C$

Explanation:

In order to find f(x) , we must take the indefinite integral of the given derivative $f'(x)=x^{2}+4x$ .

Noting that the power rule for integrals is

$\int x^{n}=\frac{x^{n+1}}{n+1}+C$ for all $n\neq-1$ and with an arbitrary constant of integration ,

we can determine that

$f(x)=\frac{1}{3}x^{3}+2x^{2}+C$ .

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Example Question #1079 : Functions

Evaluate the indefinite integral:

$\int 3xsin(x^2) dx$

Possible Answers:

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

3cos(x^2)+C

cos(x^2)+C

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

Correct answer:

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

Explanation:

To integrate, one must use the "u-substitution" or substitution method for integration, because one cannot simply integrate $\int 3xsin(x^2)dx$ .

In this case, we are calling u=x^2 , and therefore du=2xdx .

Replace x^2 with , and with $\frac{du}{2x}$ .

The integral becomes far easier:

$\frac{3}{2}\int sin(u)du$ , and is equal to $-\frac{3}{2}cos(u)+C$ .

The rule used for this integration is

$\int (sinu)du=-cosu+C$ .

(The constant -3/2 remains unchanged.) Finally, replace with x^2 to go back to the original variable/term being integrated.

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Example Question #1081 : Functions

Evaluate $\int_{1}^{4}2x+1dx$ .

Possible Answers:

Correct answer:

Explanation:

Take the anti-derivative (integrate using the Power Rule : $\int u^ndu={\frac{1}{n+1}}u^{n+1}$ ) of the expression:

$\int_{1}^{4}2x+1dx=[x^2+x]_1^4$

Plug in the number at the top of the integral:

(4)^2+4=20

Then, plug in the number at the bottom of the integral:

(1)^2+1=2

Lastly, subtract the top from the bottom:

20-2=18

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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 #21 : Equations

Example Question #21 : How To Find Integral Expressions

Example Question #21 : How To Find Integral Expressions

Example Question #1074 : Functions

Example Question #21 : How To Find Integral Expressions

Example Question #22 : How To Find Integral Expressions

Example Question #1077 : Functions

Example Question #1078 : Functions

Example Question #1079 : Functions

Example Question #1081 : Functions

All Calculus 1 Resources

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