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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #8 : Definite Integrals

Evaluate: $\int_{0}^{100}dx$

Possible Answers:

100

100+C

Correct answer:

100

Explanation:

The integral can be rewritten as:

$\int_{0}^{100}dx= \int_{0}^{100} 1\:dx$

The integration of a constant is simply the constant times the variable it is integrating with respect to. Plug in the upper bound and subtract after substituting the lower bound.

$\int_{0}^{100} 1\:dx = x|_{0}^{100} = 100-0 = 100$

There is no need to add a term at the end when we are dealing with definite integrals.

The answer is: 100

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Example Question #9 : Definite Integrals

Evaluate: $\int_{\frac{1}{2}}^{1} x-2 \:dx$

Possible Answers:

$\frac{11}{8}$

$\frac{3}{8}$

$\frac{3}{5}$

$\frac{1}{8}$

$-\frac{5}{8}$

Correct answer:

$-\frac{5}{8}$

Explanation:

Integrate each term.

$\int_{\frac{1}{2}}^{1} x-2 \:dx = \frac{x^2}{2} - 2x|_{\frac{1}{2}}^{1}$

Substitute the upper bound and subtract the quantity after substituting of the lower bound.

$\frac{x^2}{2} - 2x|_{\frac{1}{2}}^{1} = \frac{1^2}{2} - 2(1) - (\frac{(\frac{1}{2})^2}{2} - 2(\frac{1}{2}))$

The right hand side of the equation becomes:

$=\frac{1}{2} - 2- ((\frac{1}{4}\times \frac{1}{2} - 2(\frac{1}{2}))$

$=\frac{1}{2} - 2- (\frac{1}{8}- 1)$

Convert all fraction to the least common denominator.

$=\frac{4}{8} - \frac{16}{8}- (\frac{1}{8}- \frac{8}{8})$

Simplify the terms.

$=-\frac{12}{8} - (-\frac{7}{8}) = -\frac{12}{8} +\frac{7}{8} = -\frac{5}{8}$

The answer is: $-\frac{5}{8}$

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Example Question #10 : Definite Integrals

Solve the integral:

$\int_{0}^{1}xe^{2x}dx$

Possible Answers:

None of the chocies.

$= \frac{1}{4}(7e^{2}-1)$

$= \frac{1}{2}(1-7e^{2})$

$= \frac{1}{4}(7e^{2}+1)$

$= \frac{1}{2}(7e^{2}+1)$

Correct answer:

$= \frac{1}{4}(7e^{2}+1)$

Explanation:

$\int xe^{2x}dx =\int udv = uv - \int vdu$

Must choose what which term is or .

u = x , take the derivative to get du = dx

$dv = e^{2x}dx$ , integrate to get $v = \frac{e^{2x}}{2}$

Plug into parts into the equation:
$uv - \int vdu$ $= xe^{2x} - \int \frac{1}{2} e^{2x}dx$

Integrating again, while including the bounds gives:

$\Big[e^{2x}(x-\frac{1}{4})\Big]^{1}_{0} = e^{2}(\frac{7}{4}) + \frac{1}{4}$ $= \frac{1}{4}(7e^{2}+1)$

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Example Question #11 : Definite Integrals

Find the integral of this equation:

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

Possible Answers:

$\frac{97}{3}$

$\frac{100}{3}$

$\frac{103}{3}$

$\frac{450}{12}$

$\frac{425}{12}$

Correct answer:

$\frac{100}{3}$

Explanation:

Distribute the polynomial first:
(x+2)(x^2+4) = x^3 + 2x^2 + 4x + 8

Then integrate:

$\int_{0}^{2}x^3 + 2x^2 + 4x + 8 \ dx = \Big[\frac{x^4}{4} + \frac{2x^3}{3}+2x^2+8x\Big]_0 ^2$

And plug and solve:

$\frac{16}{4} + \frac{16}{3} + 8 + 16 - 0=\frac{100}{3}$

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Example Question #21 : Finding Integrals

Evaluate the integral:

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

Possible Answers:

$\frac{3}{2}$

$-\frac{3}{2}$

Correct answer:

$-\frac{3}{2}$

Explanation:

Factor the polynomial:

(x+1)(x-2) = x^2 - x -2

Then integrate (To integrate, add 1 power of each term. So $x^2\rightarrow x^3$ . Then divide the term by the value of the exponent. $x^3 \rightarrow \frac{1}{3}x^3$ .

$\int_{0}^{3} (x^2 - x -3 ) = \Big[\frac{1}{3}x^3-\frac{1}{2}x^2-2x\Big]_{0}^{3}$

Then evaulate the integral:

$[\frac{1}{3}(3)^3 - \frac{1}{2}(3)^3-2(3)] - [\frac{1}{3}(0)^3 - \frac{1}{2}(0)^3-2(0)]$

$= 9 - \frac{9}{2} - 6$

= $-\frac{3}{2}$

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Example Question #22 : Finding Integrals

$\int_{0}^{5} \frac{x+5}{x^2+x+-20} dx$

Possible Answers:

-ln(2)

ln(4)

ln(-4)

-ln(4)

ln(2)

Correct answer:

-ln(4)

Explanation:

Simplify:

$\frac{x+5}{x^2+x-20} = \frac{x+5}{(x+5)(x-4)} = \frac{1}{x-4}$

Integrate:

$\int_{0}^{5}\frac{1}{x-4}dx = \Big[ln(x-4)\Big]_{0}^{5}$

Evaluate:

= ln|(5) - 4| - ln|(0) - 4|

= -ln(4)

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Example Question #23 : Finding Integrals

Find the integral:

$\int_{0}^{5} \frac{x+2}{x^2+4x+4} dx$

Possible Answers:

$ln(\frac{7}{3})$

$ln(\frac{2}{5})$

$ln(\frac{7}{2})$

Correct answer:

$ln(\frac{7}{2})$

Explanation:

$\frac{x+2}{x^2+4x+4} = \frac{x+2}{(x+2)(x+2)} = \frac{1}{x+2}$

Integrate new equation:

$\int_{0}^{5} \frac{1}{x+2} dx = \Big[ln\left |(x+2) \right |\Big]_{0}^{5}$

$= ln\left |(5)+2\right | - ln\left |((0)+2)\right | = ln(7) - ln(2)$

Use natural log rules:

$ln(7)-ln(2) = ln(\frac{7}{2})$

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Example Question #15 : Definite Integrals

$\int_{1}^{3} \frac{x^2 +8x + 15}{x^2+3x} dx$

Evaulate the integral:

$\int_{1}^{3} \frac{x^2 +8x + 15}{x^2+3x} dx$

Possible Answers:

= 2+ 5ln(3)

= 2+ 5ln(2)

= 2 - 5ln(3)

= 4+ 5ln(3)

Correct answer:

= 2+ 5ln(3)

Explanation:

Simplify:

$\frac{x^2 +8x + 15}{x^2+3x} = \frac{(x+3)(x+5)}{x(x+3)} = \frac{x+5}{x}$

Seperate the terms:

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

Now integrate the simplified form:

$\int_{1}^{3} (1 + \frac{5}{x}) dx$

$= \Big[x + 5ln|x|\Big]_{1}^{3}$

= [3 + 5ln|3|] - [1 + 5ln|1|]

= 2+ 5ln(3)

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Example Question #24 : Finding Integrals

Evaluate the definite integral

$\int_{0}^{\pi /2} [sin(x)+cos(x)]\,dx$

Possible Answers:

$\pi/4$

$\frac{\pi}{2}$

Correct answer:

Explanation:

Because integration is a linear operation, we are able to anti-differentiate the function term by term.

We use the properties that

The anti-derivative of is
The anti-derivative of is

to solve the definite integral

$\int_{0}^{\pi /2} [sin(x)+cos(x)]\,dx=(-cos(x)+sin(x)\Big|_{0}^{\frac{\pi}{2}}$

And by the corollary of the Fundamental Theorem of Calculus

$(-cos(x)+sin(x)\Big|_{0}^{\frac{\pi}{2}}=[-cos(\frac{\pi}{2})+sin(\frac{\pi}{2})]-[-cos(0)+sin(0)]$

=[0+1]-[-1+0]=1+1=2

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Example Question #31 : Finding Integrals

Evaluate the definite integral

$\int_{1}^{4} \frac{1}{x}\,dx$

Possible Answers:

DNE

$-\frac{3}{4}$

$ln \, 4$

Correct answer:

$ln \, 4$

Explanation:

We use the property that

The antii-derivative of $\frac{1}{x}$ is

to solve the definite integral

$\int_{1}^{4} \frac{1}{x}\,dx=(ln|x|)\Big|_{1}^{4}$

And by the corollary of the Fundamental Theorem of Calculus

$(ln|x|)\Big|_{1}^{4}=[ln|4|]-[ln|1|]$

And since $ln\,1=0$

the definite integral becomes

$=ln \,4$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #8 : Definite Integrals

Example Question #9 : Definite Integrals

Example Question #10 : Definite Integrals

Example Question #11 : Definite Integrals

Example Question #21 : Finding Integrals

Example Question #22 : Finding Integrals

Example Question #23 : Finding Integrals

Example Question #15 : Definite Integrals

Example Question #24 : Finding Integrals

Example Question #31 : Finding Integrals

All Calculus 2 Resources

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