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Calculus 1 : Integral Expressions

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Example Questions

Example Question #1112 : Functions

Evaluate the following indefinitie integral:

$\int (6x^{5}+2x)dx$

Possible Answers:

$x^{6}+x^{2}$

$6x^{6}+2x^{2}$

$6x^{6}+2x^{2}+C$

$x^{6}+x^{2}+C$

Correct answer:

$x^{6}+x^{2}+C$

Explanation:

To evaluate the integral, use the inverse power rule.

$\int x^{n}dx = \frac{x^{n+1}}{n+1} n\neq1$

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

$\frac{6x^{5+1}}{5+1}=\frac{6x^{6}}{6}=x^{6}$

And the following for the second term:

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

We can combine these terms and add our "C" to get the final answer:

$x^{6}+x^{2}+C$

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Example Question #1113 : Functions

Evaluate the following indefinite integral:

$\int (4x+5)dx$

Possible Answers:

$2x^{2}+5+C$

$4x^{2}+5x+C$

$2x^{2}+5x+C$

$2x^{3}+5+C$

Correct answer:

$2x^{2}+5x+C$

Explanation:

To evaluate the integral, use the inverse power rule.

$\int x^{n}dx = \frac{x^{n+1}}{n+1} n\neq1$

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

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

And the following for the second term:

$\frac{5x^{0+1}}{0+1}=\frac{5x^{1}}{1}=5x$

We can combine these terms and add out "C" to get the final answer:

$2x^{2}+5x+C$

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Example Question #1114 : Functions

Evaluate the following indefinite integral:

$\int (2x+3)dx$

Possible Answers:

$2x^{2}+3$

$x^{2}+3x$

$x^{2}+3x+C$

$2x^{2}+3x+C$

Correct answer:

$x^{2}+3x+C$

Explanation:

To evaluate the integral, use the inverse power rule.

$\int x^{n}dx = \frac{x^{n+1}}{n+1} n\neq1$

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

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

And the following for the second term:

$\frac{3x^{0+1}}{0+1}=\frac{3x^{1}}{1}=3x$

We can combine these terms and add our "C" to get the final answer:

$x^{2}+3x+C$

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Example Question #1115 : Functions

Which of the following integrals expresses the population of a square city with side lengths of 4miles and population density given by $\rho(x)$ people/mile^2 where is distance from the north boundary of the city?

Possible Answers:

$\int 4\rho(x)dx$

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

$\int_{0}^{4} 4xdx$

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

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

Correct answer:

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

Explanation:

Here we want to integrate slices of the city with approximately uniform density. From the north side, we can divide the city into rectangles of length and width $\Delta x$ . Each rectangle has a population density of approximately $\rho(x_{k})$ , where $x_{k}$ is the distance from the north side of the $k^{th}$ rectangle. Thus, the area of each rectangle is $4\Delta x mi^{2}$ and the total population of the rectangle is approximately $4\rho(x_{k})\Delta x$ people. We sum up all of the rectangles on the interval [0,4] and take the limit as the number of rectangles on the interval approaches infinity. This limit takes the width of each rectangle and makes it infinitely small, making the density not just approximately uniform, but approaching perfectly uniform. This limit of a sum is one definition of the definite integral.

$pop \approx\sum_{k=0}^{n} 4\rho(x_{k})\Delta x$ where $x_{k}=0+k\Delta x$ and $\Delta x= \frac{4}{n}$ .

$pop=\lim_{n\to\infty} \sum_{k=0}^{n}4\rho(x_{k})\Delta x=\int_{0}^{4}4\rho(x)dx$

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Example Question #64 : Equations

Find the indefinite integral

$\int\sin (3x)-9x^2+4e^{2x}dx$ .

Possible Answers:

$\frac{-\cos (3x)}{3}-3x^3+4e^{2x}+C$

$\frac{-\cos (3x)}{3}-9x^3+2e^{2x}+C$

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

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

None of these

Correct answer:

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

Explanation:

An integral is the opposite of a derivative. For the general integration rules for this problem we need to know that the integral of x^n is $\frac{x^{n+1}}{n+1}$ .

The integral of sine is negative cosine and the integral of e^x is itself. We must also use the chain rule for sin and for the exponential function which states that the derivative of f(g(x)) is f'(g(x))g'(x) .

The integral of $\sin (3x)$ is

$\frac{-\cos (3x)}{3}$ .

The integral of -9x^2 is 3x^3 . The integral of $4e^{2x}$ is $2e^{2x}$ .

We must also add a C as the integral is indefinite.

The final answer is

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

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Example Question #2141 : Calculus

Find the indefinite integral.

$\int3x^5+7x^4+x^3+6x^2-10x-15 dx$

Possible Answers:

$\frac{x^{6}}{2}+7x^5}+\frac{x^{4}}{4}+x^3-5x^2-15x+C$

None of these

$\frac{x^{6}}{2}+\frac{7x^5}{5}+\frac{x^{4}}{4}+2x^3-5x^2-15x+C$

$\frac{x^{6}}{2}+\frac{7x^5}{5}+\frac{x^{4}}{4}+x^3-x^2-15x+C$

$\frac{x^{6}}{6}+\frac{7x^5}{5}+\frac{x^{4}}{4}+2x^3-5x^2-15x+C$

Correct answer:

$\frac{x^{6}}{2}+\frac{7x^5}{5}+\frac{x^{4}}{4}+2x^3-5x^2-15x+C$

Explanation:

An integral is the opposite of a derivative. The rule for integrating x^n is $\frac{x^{n+1}}{n+1}$ .

This is enough information to find the integral of this function.

The integral can be found to be

$\frac{x^{6}}{2}+\frac{7x^5}{5}+\frac{x^{4}}{4}+2x^3-5x^2-15x+C$

The C is the constant of integration for indefinite integrals.

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Example Question #67 : Integral Expressions

Find the definite integral.

$\int_{0}^{\frac{\pi}{2}}6\cos (3x)-12x^2 dx$

Possible Answers:

$-2-4\pi$

None of these

$-1-4\pi^3$

$-2-\frac{\pi^3}{2}$

$-4\pi^3$

Correct answer:

$-2-\frac{\pi^3}{2}$

Explanation:

An integral is the opposite of a derivative. The integration rules needed for this problem are as follows.

The integral of x^n is $\frac{x^{n+1}}{n+1}$ and the integral of cosine is sine.

This makes the integral

$2\sin (3x)-4x^3$ .

We must calculate this integral at the upper limit and subtract its value at the lower limit.

At the upper limit of $\frac{\pi}{2}$ we can calculate

$2\sin \left(\frac{3\pi}{2}\right)-4\pi^3=-2-4\left(\frac{\pi}{2}\right)^3$

At the lower limit of 0

$2\sin (3\cdot 0)-4\cdot 0^3=0$ so the final answer is

$-2-\frac{\pi^3}{2}$ .

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Example Question #68 : Integral Expressions

Find the average value of the function

$f(x)=\frac{5(x^2+5)}{x^2}$ on the interval [1,3]

Keep your answer in the form of a fraction. (An exact answer)

Possible Answers:

$\frac{42}{17}$

$\frac{23}{99}$

$\frac{40}{3}$

$\frac{2\sqrt{2}}{7}$

Correct answer:

$\frac{40}{3}$

Explanation:

$\frac{5x^2}{x^2}+\frac{25}{x^2}=5+25x^{-2}$

Our bounds are given as [1,3]. Recall the power rule for integration:

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

For an average value, find the integral then divide it by your upper bound minus your lower bound.

You get,

$\frac{\int_{1}^{3} 5+25x^{-2}dx}{3-1}\rightarrow \frac{5x-25x^{-1}}{2}$

Evaluate this expression from to . That is, plug your upper bound into the expression for the variable ( in this case). Then subtracted the value you obtain from plugging in the lower bound for from the value you got when plugging in the upper bound.

$\frac{5x-25x^{-1}}{2}\rightarrow \frac{5\cdot 3-25\cdot (3^{-1})}{2}-\frac{5\cdot1-25\cdot1^{-1}}{2}$

$=\frac{15-\frac{25}{3}}{2}-\frac{5-25}{2}=\frac{40}{3}$

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Example Question #69 : Integral Expressions

Solve the definite integral.

$\int_{1}^{5} (6x^2-18x-2) dx$

Possible Answers:

None of these

Correct answer:

Explanation:

To solve a definite integral you must first find the equation for the integral. Then you must plug in the upper limit and the lower limit and subtract the two values.

The integral of x^n is $\frac{x^{n+1}}{n+1}$ .

The integral expression equals

2x^3-9x^2-2x .

Plugging in 5 gets

2(5)^3-9(5)^2-2(5)=15 .

Plugging in 1 gets

2(1)^3-9(1)^2-2(1)=-9 .

Finally the difference gives

15-(-9)=24 .

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Example Question #66 : Equations

Find the indefinite integral.

$\int x\cos (2x)dx$

Possible Answers:

$\frac{\sin (2x)}{2}+\frac{\cos (2x)}{4}+C$

$\frac{x\sin (x)}{2}+\frac{\cos (x)}{4}+C$

$\frac{x\sin (2x)}{4}+\frac{\cos (2x)}{4}+C$

$\frac{x\sin (2x)}{2}+\frac{\cos (2x)}{4}+C$

None of these

Correct answer:

$\frac{x\sin (2x)}{2}+\frac{\cos (2x)}{4}+C$

Explanation:

For this problem we must use the strategy of integration by parts. It states that

$\int udv=uv-\int vdu$ .

For this problem we set

u=x and dv=cos (2x) .

The derivative of x^n is $nx^{n-1}$ and the integral of cos is sin.

Thus

du=1 and $v=\frac{\sin (2x)}{2}$

The answer is then

$\frac{x\sin (2x)}{2}-\int \frac{\sin (2x)}{2}dx$ .

The integral of sin is -cos so the final answer is

$\frac{x\sin (2x)}{2}+\frac{\cos (2x)}{4}+C$ .

C is the constant of integration and we must add it to all indefinite integrals.

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Calculus 1 : Integral Expressions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1112 : Functions

Example Question #1113 : Functions

Example Question #1114 : Functions

Example Question #1115 : Functions

Example Question #64 : Equations

Example Question #2141 : Calculus

Example Question #67 : Integral Expressions

Example Question #68 : Integral Expressions

Example Question #69 : Integral Expressions

Example Question #66 : Equations

All Calculus 1 Resources

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