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Calculus 2 : Integrals

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Example Question #231 : Integrals

Give the area of the region bounded by the curves of the functions

$f(x) = 4e^{x}$

and

$g(x) = e^{2x} + 3$ .

Possible Answers:

$4 - \ln 3$

$12 \ln 3$

$4 + \ln 3$

$4 + \ln 27$

$4 - \ln 27$

Correct answer:

$4 - \ln 27$

Explanation:

The two curves intersect when

g(x) = f(x) .

$e^{2x} + 3 = 4e^{x}$

$e^{2x} - 4e^{x} + 3 = 0$

$\left (e^{x} \right )^{2} - 4e^{x} + 3 = 0$

$\left ( e^{x} - 1 \right ) \left ( e^{x} - 3 \right ) = 0$

$e^{x} = 1$

x = 0

$e^{x} = 3$

$x = \ln 3$

Therefore, the area between the curves is

$\left |\int_{0}^{\ln 3} \left [g(x) - f(x) \right ] \mathrm{d}x \right |$

$= \left |\int_{0}^{\ln 3} \left (e^{2x} - 4e^{x} + 3 \right ) \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \frac{e^{2x}}{2} -4e^{x}+ 3x \\ \\ \end{matrix}\right| \begin{matrix} \ln 3\\ \\ 0 \end{matrix} \right |$

$=\left |\left ( \frac{e^{2 \ln 3}}{2} -4e^{\ln 3}+ 3 \ln 3 \right )- \left ( \frac{e^{2 \cdot 0}}{2} -4e^{0}+ 3 \cdot 0 \right ) \right |$

$=\left |\left ( \frac{e^{ \ln 9}}{2} -4e^{\ln 3}+ 3 \ln 3 \right )- \left ( \frac{e^{ 0}}{2} -4e^{0}+ 0 \right ) \right |$

$\; =\left | \left ( \frac{9}{2} - 4 \cdot 3 + \ln 27 \right) - \left ( \frac{1}{2} -4 \cdot 1 + 0 \right ) \right |$ $\; =\left | \left ( \frac{9}{2} - 4 \cdot 3 + \ln 27 \right) - \left ( \frac{1}{2} -4 \cdot 1 + 0 \right ) \right |$

$\; = \left |\left ( \frac{9}{2} -12 + \ln 27 \right) - \left ( \frac{1}{2} -4 \right ) \right |$

$=\left | \frac{9}{2} -12 + \ln 27 - \frac{1}{2} + 4 \right |$

$=\left | -4 + \ln 27 \right | = 4 - \ln 27$

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Example Question #232 : Integrals

Give the area of the region bounded by the curves of the functions

f(x) = x^3 + x - 5

and

g(x) = 17x - 5 .

Possible Answers:

128

256

Correct answer:

128

Explanation:

The two curves intersect when

f(x) = g(x) .

x^3 + x - 5 = 17x - 5

x^3 + x - 5 - 17x +5 =0

x^3 -16x =0

$x\left ( x^2 -16 \right ) =0$

$x\left ( x+4 \right )\left ( x-4 \right ) =0$

x = -4,0,4

So the curves intersect at three points: x = -4,0,4 .

The area of the region bound by the curves is

$\int_{-4}^{0}\left |f(x) - g(x) \right | \mathrm{d}x+ \int_{0}^{4}\left |f(x) - g(x) \right | \mathrm{d}x$

$=\left |\int_{-4}^{0}\left ( x^3 -16x \right )\mathrm{d}x \right |+\left | \int_{0}^{4} \left ( x^3 -16x \right ) \; \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \left ( \frac{x^{4}}{4} -\frac{16x^{2}}{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 0\\ \\ -4 \end{matrix} \right | + \left | \left.\begin{matrix} \left ( \frac{x^{4}}{4} -\frac{16x^{2}}{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 4\\ \\ 0 \end{matrix} \right |$

$=\left | \left.\begin{matrix} \left ( \frac{x^{4}}{4} -8x^{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 0\\ \\ -4 \end{matrix} \right | + \left | \left.\begin{matrix} \left \left ( \frac{x^{4}}{4} -8x^{2} \right ) \\ \\ \end{matrix}\right|\begin{matrix} 4\\ \\ 0 \end{matrix} \right |$

$=\left | \left ( \frac{0^{4}}{4} -8 \cdot 0^{2} \right ) - \left ( \frac{(-4)^{4}}{4} -8 \cdot (-4)^{2} \right ) \right | + \left | \left ( \frac{4^{4}}{4} -8 \cdot 4^{2} \right ) - \left ( \frac{0^{4}}{4} -8 \cdot 0^{2} \right ) \right |$

$=\left | 0 - \left ( \frac{256}{4} -8 \cdot 16 \right ) \right | + \left | \left ( \frac{256}{4} -8 \cdot 16 \right ) - 0 \right |$

$=\left | - \left (64 -128\right ) \right | + \left | \left ( 64 -128 \right ) \right |$

= 64 + 64 = 128

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Example Question #233 : Integrals

Give the area of the region bounded by the curves of the functions

$f(x) = \frac{16}{x}$

and

g(x) = 10 -x .

Possible Answers:

$16 \ln 4$

$30 - 16 \ln 4$

$30 - 16 \ln 16$

$56 - 16 \ln 16$

$56 - 16 \ln 4$

Correct answer:

$30 - 16 \ln 4$

Explanation:

The two curves intersect when

g(x) = f(x) .

$\frac{16}{x} =10 -x$

$x - 10 + \frac{16}{x} = 0$

Since it can be assumed that $x \neq 0$ ,

$x \left ( x - 10 + \frac{16}{x} \right )=x\cdot 0$

$x ^{2}- 10x + 16 = 0$

(x-2) (x-8)= 0

Therefore, either x-2 = 0 , in which case x = 2 , or x - 8 = 0 , in which case x = 8 .

Therefore, the area between the curves is

$\int_{2}^{8}\left |g(x) - f(x) \right | \mathrm{d}x$

$=\left |\int_{2}^{8} \left ( x - 10 + \frac{16}{x} \right ) \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \frac{x^2}{2} -10x + 16 \ln x\\ \\ \end{matrix}\right| \begin{matrix} 8\\ \\ 2 \end{matrix} \right |$

$=\left | \left (\frac{8^2}{2} -10 \cdot 8 + 16 \ln 8 \right ) - \left ( \frac{2^2}{2} -10 \cdot 2 + 16 \ln 2 \right ) \right |$

$=\left | \left (32-80 + 16 \ln 8 \right ) - \left ( 2-20 + 16 \ln 2 \right ) \right |$

$=\left | \left (-48 + 16 \ln 8 \right ) - \left ( -18 + 16 \ln 2 \right ) \right |$

$=\left | -48 + 18 + 16 \ln 8 - 16 \ln 2 \right |$

$= \left | -30 + 16( \ln 8 - \ln 2 ) \right | =\left | -30 + 16 \ln 4 \right |= 30 - 16 \ln 4$

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Example Question #1 : Area Under A Curve

Find the area under the curve f(x) = e^x from x = -2 to x = 2 .

Possible Answers:

$e^2 + 2e^{-2}$

$1-2e^{-2}$

2e^2

$e^2 - e^{-2}$

2e^2 - 1

Correct answer:

$e^2 - e^{-2}$

Explanation:

The area under a curve f(x) between two x-values $x_{1},x_2$ corresponds to the integral

$\int_{x_1}^{x_2} f(x)dx$ .

In this case, f(x) = e^x, x_1 = -2, x_2 = 2 , and we can find by the fundamental theorem of calculus that...

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

= [e^x] evaluated from to

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

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Example Question #2 : Area Under A Curve

Find the area between f(x) and g(x) , such that $f(x) = \sqrt{x}$ and g(x) = x^3

Possible Answers:

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{5}{12}$

$\frac{1}{3}$

$\frac{7}{12}$

Correct answer:

$\frac{5}{12}$

Explanation:

The first step to find the area between two curves is to set them equal to each other and find their intersection points.

$\\f(x) = g(x)\\ \sqrt{x} = x^3\\ x = x^6\\ x = 0, x = 1$

The next step is to find which curve is "on the top" and which is "on the bottom" inside the interval (0,1) . We find that f(x) is on the top and g(x) is on the bottom.

Now, to find the area between the two curves, we take the integral from to of the top curve minus the bottom curve.

$\\\int_{0}^{1}(f(x) - g(x))dx\\ \int_{0}^{1}(\sqrt{x} - x^3)dx\\ \left [ \frac{2}{3}x^{3/2} - \frac{1}{4}x^4 \right ]$ evaluated from to

$\\= \left(\frac{2}{3} - \frac{1}{4}\right) - (0 - 0)\\ = \frac{5}{12}$

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Example Question #236 : Integrals

Find the total area under the curve $f(x)=2xe^{x^2}$ from $x = 0\ to\ x = 5.$

Possible Answers:

$\frac{25e^{25}}{2}$

$10e^{25}$

$e^{25}$

$e^{25}-1$

Correct answer:

$e^{25}-1$

Explanation:

The definite integral of f(x) over the interval [0, 5] is the area under the curve.

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

A substitution makes this integral easier to evaluate. Let $u=e^{x^2}$ . Then $du=2xe^{x^2}dx$ . We can also change the limits of integration so that they are in terms of u. When x=5, $u=e^{25}$ . When x=0, $u=e^{0}=1$ . Making these substitutions, the integral becomes

$\int_{1}^{e^{25}}du=u \left |_{1}^{25}=e^{25}-1$

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Example Question #1 : Area Under A Curve

What is the area under the curve $f(x)=\frac{x}{x^2+1}$ over the interval $1\le x\le 2$ ?

Possible Answers:

$\frac{1}{2}ln\left(\frac{5}{2}\right)$

$\frac{ln5}{2}$

$\frac{21}{5}$

e^2-1

Correct answer:

$\frac{1}{2}ln\left(\frac{5}{2}\right)$

Explanation:

The area under a curve f(x) over the interval [a, b] is $\int_{a}^{b}f(x)dx$ .

In this example, this leads to the definite integral

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

A substitution makes the antiderivative of this function more obvious.

Let

$u=x^2+1,\ du=2xdx, \frac{du}{2}=xdx.$ .

We can also convert the limits of integration to be in terms of to simplify evaluation. When $x=1,\ u=2$ , and when $x=2, \ u=5$ .

Making these substitutions results in

$\int_{2}^{5}\frac{\frac{du}{2}}{u}=\frac{1}{2}\int_{2}^{5}\frac{du}{u}$ .

Recall that in general $\int\frac{dx}{x}=ln(x)+c$ , so evaluating the integral leads to

$\frac{1}{2}\int_{2}^{5}\frac{du}{u}=\frac{1}{2}ln(u)|_{2}^{5}=\frac{1}{2}(ln(5)-ln(2))$

$=\frac{1}{2}ln\left(\frac{5}{2}\right)$

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Example Question #238 : Integrals

Find the following indefinite integral:

$\int \frac{1}{x^2+x-6} \ dx$

Possible Answers:

$\ln|x^2+x-6|+C$

$\frac{1}{5}}\ln|x-2|-\frac{1}{5}\ln|x+3|$

$-\frac{1}{5}}\ln|x-2|+\frac{1}{5}\ln|x+3| + C$

$\ln|x^2+x-6|$

$\frac{1}{5}}\ln|x-2|-\frac{1}{5}\ln|x+3| + C$

Correct answer:

$\frac{1}{5}}\ln|x-2|-\frac{1}{5}\ln|x+3| + C$

Explanation:

In order to find the anti-derivative, lets first factor the denominator.

$\int \frac{1}{x^2+x-6} \ dx$

$= \int \frac{1}{(x+3)(x-2)} \ dx$

Now we use partial-fraction decomposition to get this into a form we know how to take the anti-derivative.

$\int \frac{A}{(x+3)} +\frac{B}{(x-2)} \ dx$

Now we need to find the values of A and B. First lets write out the equation.

1=A(x-2)+B(x+3)

1=Ax-2A+Bx+3B

1=x(A+B)+(3B-2A)

Now we can get our system of equations in order to find A and B.

$\\ 1: A+B=0 \\ 2: 3B-2A=1$

From equation 1, we get

A=-B

Now we can substitute this into equation 2.

3B-2(-B)=1

$5B=1 \rightarrow B=\frac{1}{5}$

Now we substitue B back into equation 1.

$A=-\frac{1}{5}$

Now we can put the values of A, and B into our problem.

$\int \frac{-\frac{1}{5}}{(x+3)} +\frac{\frac{1}{5}}{(x-2)} \ dx$

Now we can simply take the anti-derivative.

$\int \frac{-\frac{1}{5}}{(x+3)} +\frac{\frac{1}{5}}{(x-2)} \ dx$

$=\frac{1}{5}}\ln|x-2|-\frac{1}{5}\ln|x+3| + C$

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Example Question #239 : Integrals

Evaluate:

$\int_{0}^{1} t \ dt$

Possible Answers:

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

To evaluate the integral, we use inverse power law.

Remember that the inverse power law is

$\int x^a =\frac{1}{a+1}x^{a+1}$ .

Lets apply this to our problem.

$\int_{0}^{1} t \ dt$

$= \frac{1}{2}t^2\left |_{0}^{1}$

From here we plug in the bounds and subtract the lower bound function value from the upper bound function value.

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

$= \frac{1}{2}-0=\frac{1}{2}$

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Example Question #240 : Integrals

Find the area under the curve of the function y = x^2 - 2x between x = 2 and x = 4 .

Possible Answers:

$\frac{25}{3}$

$\frac{20}{3}$

$\frac{32}{3}$

$\frac{64}{3}$

Correct answer:

$\frac{20}{3}$

Explanation:

Asking for the area under the curve means you are going to be doing an integral of the function provided. The bounds are given for you. The integral thus looks like this: $\int_{2}^{4} x^{2} - 2x dx$ .

When integrated, you get

$\frac{x^3}{3} - x^2$ evaluated at x = 4 and x = 2 .

At x = 4 the expression evaluates to $\frac{64}{3} - 16$ and at x = 2 it evalutes to $\frac{8}{3} - 4$ .

Subtracting the two gives you your final answer of $\frac{20}{3}$ .

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #231 : Integrals

Example Question #232 : Integrals

Example Question #233 : Integrals

Example Question #1 : Area Under A Curve

Example Question #2 : Area Under A Curve

Example Question #236 : Integrals

Example Question #1 : Area Under A Curve

Example Question #238 : Integrals

Example Question #239 : Integrals

Example Question #240 : Integrals

All Calculus 2 Resources

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