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Calculus 2 : Finding Integrals

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Example Question #27 : Definite Integrals

Find the area between y=sin(x) and y=sec^2(x) between $x:(0, \frac{\pi}{4})$

Possible Answers:

$\frac{\sqrt2}{2}$

$\frac{\sqrt2}{2}+2$

Correct answer:

$\frac{\sqrt2}{2}$

Explanation:

We can write this problem as:

$\int_{0}^{\frac{\pi}{4}}(sec^2(x)-sin(x))dx$

Integrating:

$\int(sec^2(x)-sin(x))dx=tan(x)+cos(x)+C=F(x)$

By the fundamental theorem of calculus:

$\int_{0}^{\frac{\pi}{4}}(sec^2(x)-sin(x))dx=F(\frac{\pi}{4})-F(0)$

$=[tan(\frac{\pi}{4})+cos(\frac{\pi}{4})]-[tan(0)+cos(0)]=[1+\frac{\sqrt2}{2}]-[1]$

$\int_{0}^{\frac{\pi}{4}}(sec^2(x)-sin(x))dx=\frac{\sqrt2}{2}$

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Example Question #21 : Definite Integrals

$Evaluate:\int_{0}^{5}(x^2+x)dx$

Possible Answers:

$\frac{175}{6}$

$\frac{200}{6}$

$\frac{325}{6}$

$\frac{325}{2}$

$\frac{250}{3}$

Correct answer:

$\frac{325}{6}$

Explanation:

Compute the Indefinite Integral

$\int (x^2+x)dx=\frac{x^3}{3}+\frac{x^2}{2}$

Evaluate the integral

$(\frac{5^3}{3}+\frac{5^2}{2})-(\frac{0^3}{3}+\frac{0^2}{2})$

$(\frac{125}{3}+\frac{25}{2})-(0+0)$

$\frac{250}{6}+\frac{75}{6}=\frac{325}{6}$

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Example Question #29 : Definite Integrals

Suppose u=ax , where is a constant

Find such that $\int_{0}^{2} 5u dx =2$

Possible Answers:

u=x

u=5x

$u=\frac{1}{5}x$

$u=\frac{5}{2}x$

Correct answer:

$u=\frac{1}{5}x$

Explanation:

$\int 5ax dx= \frac{5}{2}ax^2 +C= F(x)$

By the fundamental theorem of calculus:

$\int_{0}^{2} 5u dx =F(2)-F(0)=2$

$\frac{5}{2}a*2^2-0=2$

10a=2

$a=\frac{1}{5}$

$u=\frac{1}{5}x$

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Example Question #30 : Definite Integrals

Evaluate this integral.

$\int_{2}^{0} 3+4x dx$

Possible Answers:

Answer not listed

Correct answer:

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) = 3 + 4x .

The antiderivative is F(x) = 3x + 2x^2 .

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

$\int_{2}^{0} 3 + 4x dx = \left (3x + 2x^2 \right ) _{2}^{0}$

$= 3(0) + 2(0)^2 - \left (3(2) + 2(2)^2 \right )$

= - 14

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Example Question #31 : Definite Integrals

Evaluate this integral.

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

Possible Answers:

-6.48

4.67

9.43

Answer not listed

2.16

Correct answer:

4.67

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) = e^x .

The antiderivative is F(x) = e^x .

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

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

$= e^{2} - e^{1}$

= 4.67

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Example Question #32 : Definite Integrals

Evaluate this integral.

$\int_{1}^{2} 16x^3 - \frac{1}{x} dx$

Possible Answers:

64.8

98.4

12.5

59.3

Answer not listed

Correct answer:

59.3

Explanation:

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

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

In this case, $f(x) = 16x^3 - \frac{1}{x}$ .

The antiderivative is F(x) =4x^4 - ln(x) .

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

$\int_{1}^{2} 16x^3 - \frac{1}{x} dx = \left (4x^4 - ln(x) \right ) _{1}^{2}$

$= 4(2)^4 - ln(2) - \left (4(1)^4 - ln(1) \right )$

= 59.3

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Example Question #33 : Definite Integrals

Evaluate this integral.

$\int_{0}^{2} x^2 dx$

Possible Answers:

$\frac{4}{5}$

$\frac{7}{8}$

Answer not listed

$\frac{8}{3}$

$\frac{3}{2}$

Correct answer:

$\frac{8}{3}$

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) = x^2 .

The antiderivative is $F(x) = \frac{1}{3}x^3$ .

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

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

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

$= \frac{8}{3}$

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Example Question #2141 : Calculus Ii

Evaluate this integral.

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

Possible Answers:

Answer not listed

Correct answer:

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) = x-5 .

The antiderivative is $F(x) = \frac{1}{2}x^2 - 5x$ .

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

$\int_{0}^{2} x-5 dx = \left ( \frac{1}{2}x^2 - 5x \right )_{0}^{2}$

$= \frac{1}{2}(2)^2 - 5(2) - \left (\frac{1}{2}(0)^2 - 5(0) \right )$

= -8

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Example Question #35 : Definite Integrals

Evaluate this integral.

$\int_{1}^{2} 12x - 1 \ dx$

Possible Answers:

Answer not listed

Correct answer:

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) = 12x -1 .

The antiderivative is F(x) = 6x^2 - x .

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

$\int_{1}^{2} x-5 \ dx = \left ( 6x^2 - x \right )_{1}^{2}$

$= 6(2)^2 - (2) - \left ( 6(1)^2 - (1)\right )$

= 17

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Example Question #36 : Definite Integrals

Evaluate this integral.

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

Possible Answers:

$\pi$

Answer not listed

Correct answer:

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:

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

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

= 2

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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #27 : Definite Integrals

Example Question #21 : Definite Integrals

Example Question #29 : Definite Integrals

Example Question #30 : Definite Integrals

Example Question #31 : Definite Integrals

Example Question #32 : Definite Integrals

Example Question #33 : Definite Integrals

Example Question #2141 : Calculus Ii

Example Question #35 : Definite Integrals

Example Question #36 : Definite Integrals

All Calculus 2 Resources

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