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Calculus 2 : Definite Integrals

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Example Questions

Example Question #41 : Definite Integrals

Evaluate this integral.

$\int_{0}^{1} \sin x dx$

Possible Answers:

5.67

Answer not listed

3.25

2.51

1.54

Correct answer:

1.54

Explanation:

$\int_{a}^{b} f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

In this case, $f(x) = \sin x$ .

The antiderivative is $F(x) = - \cos x$ .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{0}^{1} \sin x \ dx = \left ( -\cos x )_{0}^{1}$

$= - \cos(1) - \left ( - \cos (0) \right )$

= 1.54

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Example Question #42 : Definite Integrals

Evaluate the following definite integral:

$\int_{-3}^{3}(1+x^2)dx$

Possible Answers:

Correct answer:

Explanation:

By a straightforward application of antiderivative rules for polynomial functions,

$\int_{-3}^{3}(1+x^2)dx$ $= \left.\ x+\frac{1}{3}x^3\right|_-_3^3$

=(3+9)-(-3-9)

=12+3+9

=24

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Example Question #43 : Definite Integrals

Evaluate the integral $\int\sec^3(x)dx$ .

Possible Answers:

$\frac{1}{2}\biggr(\sec(x)\tan(x)+\ln\big|\sec(x)-\tan(x)\big|\biggr)+C$

$-\frac{1}{2}\biggr(\sec(x)\tan(x)+\ln\big|\sec(x)+\tan(x)\big|\biggr)+C$

$\frac{1}{2}\biggr(\sec(x)\tan(x)-\ln\big|\sec(x)-\tan(x)\big|\biggr)+C$

$\frac{1}{2}\biggr(\sec(x)\tan(x)+\ln\big|\sec(x)+\tan(x)\big|\biggr)+C$

$\frac{1}{2}\biggr(\sec(x)\tan(x)-\ln\big|\sec(x)+\tan(x)\big|\biggr)+C$

Correct answer:

$\frac{1}{2}\biggr(\sec(x)\tan(x)+\ln\big|\sec(x)+\tan(x)\big|\biggr)+C$

Explanation:

This integral is tricky, because there doesn't seem to be an apparent substitution method to use. The derivative of $\sec(x)$ , $\sec^2(x)$ , and $\sec^3(x)$ seem to only complicate the integral further. However, we do note that the integral of $\sec^2(x)$ is given by $\tan(x)$ . Thus, we can use integration by parts here. Splitting $\int\sec^3(x)dx$ into $\int\big(\sec(x)\big)\big(\sec^2(x)dx\big)$ and letting $u=\sec(x)$ and dv=sec^2(x)dx , this gives us $du=\sec(x)\tan(x)dx$ and $v=\tan(x)$ . Plugging this into the integration by parts formula, we get:

$\int\sec^3(x)dx=\int\big(\sec(x)\big)\big(\sec^2(x)dx\big)=\int udv=uv-\int vdu$

$\int\sec^3(x)dx=\big(\sec(x)\big)\big(\tan(x)\big)-\int\big(\tan(x)\big)\big(\sec(x)\tan(x)dx\big)$

$\int\sec^3(x)dx=\sec(x)\tan(x)-\int\sec(x)\tan^2(x)dx$

Unfortunately, this still looks more complicated than the original. However, using the trig identity $\tan^2(x)=\sec^2(x)-1$ on the second integral will give us something we can work with. Doing so we obtain:

$\int\sec^3(x)dx=\sec(x)\tan(x)-\int\sec(x)\big(\sec^2(x)-1\big)dx$

$\int\sec^3(x)dx=\sec(x)\tan(x)-\int\big(\sec^3(x)-\sec(x)\big)dx$

$\int\sec^3(x)dx=\sec(x)\tan(x)-\int\sec^3(x)dx+\int\sec(x)dx$

Now here, we know how to evaluate the integral of $\sec(x)$ , and we also have a copy of our integral $\int\sec^3(x)dx$ subtracted on the right side. Therefore, we can add this integral to each side of the equation, eliminating it on the right side and leaving something we can evaluate. Doing so, we obtain:

$2\int\sec^3(x)dx=\sec(x)\tan(x)+\int\sec(x)dx$

$2\int\sec^3(x)dx=\sec(x)\tan(x)+\ln\big(\sec(x)+\tan(x)\big)+D$

Finally, dividing the whole equation by and replacing the $\frac{D}{2}$ constant with a new constant gives us our final result:

$\int\sec^3(x)dx=\frac{1}{2}\biggr(\sec(x)\tan(x)+\ln\big(\sec(x)+\tan(x)\big)\biggr)+C$

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Example Question #43 : Definite Integrals

Evaluate.

$\int_{0}^{1} x^4 \ dx$

Possible Answers:

$\frac{4}{5}$

$\frac{1}{5}$

$\frac{2}{7}$

Answer not listed

$\frac{3}{5}$

Correct answer:

$\frac{1}{5}$

Explanation:

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{0}^{1} x^4 \ dx = \left ( \frac{1}{5}x^5 \right )_{0}^{1}$

$= \frac{1}{5}(1)^5 - \left ( \frac{1}{5}(0)^5 \right )$

$= \frac{1}{5}$

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Example Question #44 : Definite Integrals

$\int_{-2}^{4}e^xdx$

Possible Answers:

e^2

$e^4-{e^2}$

e^6

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

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

Correct answer:

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

Explanation:

The antiderivative of e^x is e^x . Now, all we need to do is plug in our bounds, remembering it is always top - bottom bound.

$\int_{-2}^{4}e^xdx=e^4-\frac{1}{e^2}.$ In the last step, I simply rewrote $-e^{-2}$ into the denominator, getting rid of the negative exponent.

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Example Question #41 : Definite Integrals

Evaluate.

$\int_{0}^{\pi} \cos(2x ) \ dx$

Possible Answers:

Answer not listed

$\pi$

Correct answer:

Explanation:

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{0}^{\pi} \cos(2x ) \ dx = \left ( \cos(2x) \right )_{0}^{\pi}$

$= \cos(2(\pi)) - \left ( \cos(2(0)) \right )$

= 0

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Example Question #42 : Definite Integrals

Evaluate.

$\int_{2}^{6} \sqrt{x-2} \ dx$

Possible Answers:

$\frac{4}{7}$

$\frac{11}{5}$

Answer not listed

$\frac{8}{9}$

$\frac{16}{3}$

Correct answer:

$\frac{16}{3}$

Explanation:

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{2}^{6} \sqrt{x-2} \ dx = \left ( \frac{2(x-2)^{\frac{3}{2}}}{3} \right )_{2}^{6}$

$= \frac{2(6-2)^{\frac{3}{2}}}{3} - \left ( \frac{2(2-2)^{\frac{3}{2}}}{3} \right )$

$= \frac{16}{3}$

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Example Question #61 : Finding Integrals

Evaluate.

$\int_{1}^{2}e^{3x-1} \ dx$

Possible Answers:

Answer not listed

65.21

23.54

-124.36

141.02

Correct answer:

141.02

Explanation:

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{1}^{2}e^{3x-1} \ dx = \left ( \frac{e^{3x-1}}{3} \right )_{1}^{2}$

$= \frac{e^{3(2)-1}}{3} - \left ( \frac{e^{3(1)-1}}{3} \right )$

= 141.02

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Example Question #43 : Definite Integrals

Evaluate.

$\int_{1}^{2} 2x + \frac{1}{x} \ dx$

Possible Answers:

15.67

Answer not listed

8.75

2.14

3.69

Correct answer:

3.69

Explanation:

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{1}^{2} 2x + \frac{1}{x} \ dx = \left ( x^2 + \ln(x) \right )_{1}^{2}$

$= 2^2 + \ln(2) - \left ( 1^2 + \ln(1) \right )$

= 3.69

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Example Question #2163 : Calculus Ii

Evaluate.

$\int_{0}^{1} \cos (2x-3) \ dx$

Possible Answers:

-0.35

1.47

1.23

2.94

Answer not listed

Correct answer:

-0.35

Explanation:

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{0}^{1} \cos (2x-3) \ dx = \left ( \frac{\sin(2x-3)}{2} \right )_{0}^{1}$

$= \frac{\sin(2(1)-3)}{2} - \left ( \frac{\sin(2(0)-3)}{2} \right )$

= -0.35

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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Definite Integrals

Example Question #42 : Definite Integrals

Example Question #43 : Definite Integrals

Example Question #43 : Definite Integrals

Example Question #44 : Definite Integrals

Example Question #41 : Definite Integrals

Example Question #42 : Definite Integrals

Example Question #61 : Finding Integrals

Example Question #43 : Definite Integrals

Example Question #2163 : Calculus Ii

All Calculus 2 Resources

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