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Calculus 1 : Functions

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Example Questions

Example Question #132 : Integral Expressions

Evaluate Integral

Evaluate,

$\int _{-2}^2\:\:e^{6x+7}dx$

Possible Answers:

$\frac{e^{24}-1}{6e^5}$

None of the above

$\frac{e^{24}+1}{6e^5}$

$\frac{e^{5}-1}{6e^{24}}$

$\frac{e^{24}-5}{6e}$

Correct answer:

$\frac{e^{24}-1}{6e^5}$

Explanation:

Compute the indefinite integral, $\int \:e^{6x+7}dx=\frac{e^{6x+7}}{6}+C$

Compute the boundaries, $\int _{-2}^2e^{6x+7}dx=\frac{e^{19}}{6}-\frac{1}{6e^5}$

Simplify,

$\frac{e^{24}-1}{6e^5}$

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Example Question #2221 : Calculus

Evaluate the following integral:

$\int( e^{2x}+\cos(2x))dx$

Possible Answers:

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

$(e^{2x}-\sin(2x))+C$

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

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

Correct answer:

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

Explanation:

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

u=2x, du=2dx

Now, rewrite the integral and integrate:

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

The integration was performed using the following rules:

$\int e^xdx=e^x+C, \int \cos(x)dx=\sin(x)=C$

Finally, replace u with our original x term:

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

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Example Question #141 : Equations

Evaluate the following integral:

$\int (x^2+7x+\sin(x))dx$

Possible Answers:

$\frac{x^3}{3}+\frac{7x^2}{2}+\cos(x)+C$

$x^3+\frac{7x^2}{2}-\cos(x)+C$

$\frac{x^3}{3}+{7x^2}-\cos(x)+C$

$\frac{x^3}{3}+\frac{7x^2}{2}-\cos(x)+C$

Correct answer:

$\frac{x^3}{3}+\frac{7x^2}{2}-\cos(x)+C$

Explanation:

The integral is equal to

$\frac{x^3}{3}+\frac{7x^2}{2}-\cos(x)+C$

and was evaluated using the following rules:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int \sin(x)dx=-\cos(x)+C$

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Example Question #142 : Equations

Integral

If f(x) is even,

$\int_{-a}^{a}f(x)dx= 2\int_{0}^{a}f(x)dx$

Possible Answers:

It is called differentiating even functions

It is called differentiating odd functions

The statement is false

The statement is true

None of the above

Correct answer:

The statement is true

Explanation:

According to the integration of even function rule,

$\int_{-a}^{a}f(x)dx= 2\int_{0}^{a}f(x)dx$

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Example Question #143 : Equations

Write the integral expression for the area under the following curve from x=1 to x=6 .

f(x)=4x^2-3x+1

Possible Answers:

$\int_{1}^{6}(8x-3)dx$

$\int_{6}^{1}(4x^2-3x+1)dx$

$\int_{1}^{6}(4x^2-3x+1)dx$

$\int_{1}^{6}4x^2-3x+1$

Correct answer:

$\int_{1}^{6}(4x^2-3x+1)dx$

Explanation:

To write the integral expression, simply set the bounds given with the smaller on the bottom and don't forget the . Thus,

$\int_{1}^{6}(4x^2-3x+1)dx$

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Example Question #144 : Equations

Find the integral of the following equation from to {-2} .

f(x)=x^2-5x-100

Possible Answers:

$\int_{2}^{-2}x^2-5x-100$

$\int_{2}^{-2}(2x-5)dx$

$\int_{2}^{-2}(x^2-5x-100)dx$

$\int_{-2}^{2}(x^2-5x-100)dx$

Correct answer:

$\int_{2}^{-2}(x^2-5x-100)dx$

Explanation:

To solve, simply plug in your bounds and equation. Make sure to note how the order of bounds was said in the problem. Thus,

$\int_{2}^{-2}(x^2-5x-100)dx$

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Example Question #144 : Integral Expressions

Evaluate the following integral:

$\int 3\sin(x)e^{\cos(x)}dx$

Possible Answers:

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

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

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

$-3e^{\sin(x)}+C$

Correct answer:

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

Explanation:

To evaluate the integral, we must perform the following subsitution:

$u=\cos(x), du=-\sin(x)dx$

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Now, rewrite the integral and integrate:

$-3\int e^udu=-3e^u+C$

The integral was found using the following rule:

$\int e^udu=e^u+C$

Finally, replace u with the original term:

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

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Example Question #145 : Equations

Evaluate: $\int_{0}^{3} 6x^2 + 5x - 1 \,\mathrm{d}x$

Possible Answers:

73.5

$\frac{37}{3}$

Correct answer:

73.5

Explanation:

To find the integral, first find the antiderivative, then evaluate at the limits of integration to find the definite integral.

Here, the antiderivative is $\small \small F(x) = 2x^3 + \frac{5}{2}x^2 - x+C$ .

$\small F(3) - F(1) = 73.5 - 0 = 73.5$

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Example Question #145 : Equations

Integrate:

$\int 3x^2+4x-1 dx$

Possible Answers:

x^3+x^2-x+C

x^3+2x^2-1

2x^3+2x^2-1+C

x^3+2x^2-x+C

x^3+2x^2+C

Correct answer:

x^3+2x^2-x+C

Explanation:

When integrating this expression, tackle each term separately. For $3x^{2}$ , leave the out and just integrate the term. Raise the exponent by and then put that result on the denominator: $3(\frac{x^3}{3})$ . Do that for each term. becomes $4(\frac{x^2}{2})$ and becomes . Put those all together and add a "C" at the end because it is an indefinite integral: x^3+2x^2-1+C .

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Example Question #145 : Integral Expressions

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

Possible Answers:

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

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

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

x^3+x^2-x+C

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

Correct answer:

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

Explanation:

Before integrating this expression, I would chop it up into three separate terms and simplify since there is only one term on the denominator: $\int x^2+2x-1dx$ . Then, integrate each term separately. When you integrate, you raise the exponent by and then put that result on the denominator. Therefore, the resulting expression after integrating is: $\frac{x^3}{3}+\frac{2x^2}{2}-x$ . Simplify and add a "C" at the end (it's an indefinite integral) and the answer is:

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

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #132 : Integral Expressions

Example Question #2221 : Calculus

Example Question #141 : Equations

Example Question #142 : Equations

Example Question #143 : Equations

Example Question #144 : Equations

Example Question #144 : Integral Expressions

Example Question #145 : Equations

Example Question #145 : Equations

Example Question #145 : Integral Expressions

All Calculus 1 Resources

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