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Integrate $f(x)=\frac {x^2}{\sqrt{x}}$ from .

$\frac {62}{5}$

$\frac {60}{7}$

$\frac {-62}{5}$

None of the Above

Explanation

Step 1: Rewrite the denominator as x to a certain power. A square root means that the exponent will have a value of $\frac {1}{2}$ .

We get $x^{\frac {1}{2}}$ .

Step 2: Divide the numerator and denominator of the function. When we divide terms with exponents, we must make sure that the bases are the same. Both bases are , so let's continue. When you divide exponents, you are subtracting them. You subtract the bottom exponent FROM the top exponent.
We get $2-\frac {1}{2}$ . We will convert the into a fraction with denominator . By default, the denominator is . We will multiply the numerator and denominator by 2, so the new fraction is $\frac {4}{2}$ .

Step 3: Subtract the converted fraction (exponent of top) and the exponent of the bottom.

$\frac {4}{2}-\frac {1}{2}=\frac {4-1}{2}=\frac {3}{2}$ .

The exponent of the term is $\frac {3}{2}$ .

Step 4: Integrate...
When we Integrate, we add to the exponent. We then divide the new function with a new exponent by that new exponent.

So, we get: $\frac {x^{\frac {3}{2}+1}}{\frac {3}{2}+1}$ .

Evaluate $\frac {3}{2}+1$ and then rewrite..
$\frac {3}{2}+1=\frac {3}{2}+\frac {2}{2}=\frac {3+2}{2}=\frac {5}{2}$

We get: $\frac {x^{\frac {5}{2}}}{\frac {5}{2}}$

Step 5: Evaluate the upper and the lower limit:

$\frac {4^{\frac {5}{2}}}{\frac {5}{2}}=\frac {\sqrt{4^5}}{\frac {5}{2}}=\frac {\sqrt{1024}}{\frac {5}{2}}=\frac {32}{\frac {5}{2}}$
Flip the denominator and multiply:

$32 \cdot \frac {2}{5}=\frac {64}{5}$

If
$\frac{1^\frac{5}{2}}{\frac{5}{2}}=\frac{1}{\frac{5}{2}}=\frac{1}{1}\times\frac{2}{5}=\frac{2}{5}$

Step 6:

Subtract lower limit from the upper limit:

$\frac{64}{5}-\frac{2}{5}=\frac{62}{5}$

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

$\frac{175}{12}$

$\frac{179}{12}$

$\frac{75}{12}$

$\frac{175}{8}$

$\frac{173}{12}$

Explanation

First, integrate. Remember to raise the exponent by 1 and then also put that result on the denominator:

$\frac{x^4}{4}+\frac{4x^3}{3}+\frac{x^2}{2}$

Next, evaluate at 2 and then 1. Subtract the results:
$(4+\frac{32}{3}+2)-(\frac{1}{4}+\frac{4}{3}+\frac{1}{2})$

Simplify:

$(6+\frac{32}{3})-(\frac{50}{24})=\frac{175}{12}$

$\begin{align*}&\text{Evaluate the indefinite integral: }\&\int(\frac{e^{(9x)}}{3}+\frac{ (64x^{10})}{7})dx\end{align*}$

$\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}+C$

$3e^{(9x)} +\frac{ (6400x^{11})}{77}$

$3e^{(9x)} +\frac{ (6400x^{11})}{77}+C$

$\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}$

Explanation

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\&\text{use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside. Integrals, in}\&\text{similar fashion, entail a division. Now, although we have two}\&\text{functions, since they're being added, we integrate separately}\&\text{and combine the results. This is known as superposition.}\&\text{Utilizing integral rules:}\&\int[e^{ax}]=\frac{e^{ax}}{a}\&\int x^{a}=\frac{x^{a+1}}{a+1}\&\text{The }a\text{ value for }\frac{e^{(9x)}}{3}\text{ is }9\&\text{The }a\text{ value for }\frac{(64x^{10})}{7}\text{ is }10\&\int(\frac{e^{(9x)}}{3}+\frac{ (64x^{10})}{7})dx=\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}+C\&\text{Now you'll note after integration we add a constant, C, because}\&\text{we're working with an indefinite integral.}\&\end{align*}$

What is the area under the curve bounded by the x-axis from x=4 to x=5?

$\frac{17}{6}$

$\frac{19}{6}$

$\frac{21}{2}$

$\frac{17}{3}$

Explanation

First, set up the integral expression: $\int_{4}^{5}x^2-3x-4dx$ . Then, integrate. Remember, when integrating, raise the exponent by 1 and then put that result on the denominator: $\frac{x^3}{3}-\frac{3x^2}{2}-4x$ . Then, evaluate at 5 and then 4. Subtract those results: $(\frac{125}{3}-\frac{75}{2}-20)-(\frac{64}{3}-\frac{48}{2}-16)$ . Simplify to get your final answer of $\frac{17}{6}$ .

Find the volume of the solid generated by rotating about the y-axis the region under the curve $y =\sqrt{x}$ , from to .

$\frac{128}{5}\pi$

$\frac{64}{3}\pi$

$\frac{\pi}{2}$

$\frac{256}{7}\pi^2$

None of the other answers

Explanation

Since we are revolving a function of around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

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

$=\int_{0}^{4}2\pi x^{3/2}dx$

$=[2\pi x^{5/2}]^{4}_{0}$

$=\frac{128}{5}\pi$ .

Evaluate $\int_{1}^{\infty}\frac{3}{x^{2}}dx$ .

$\infty$

Explanation

By the Formula Rule, we know that $\int_{a}^{\infty}F(x)dx=\lim_{b\rightarrow \infty}\int_{a}^{b}f(x)dx$ . We therefore know that $\int_{1}^{\infty}\frac{3}{x^{2}}dx=\lim_{b\rightarrow \infty}\int_{1}^{b}\frac{3}{x^{2}}dx$ .

Continuing the calculation:

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

$=3\int_{1}^{b}\frac{1}{x^{2}}dx$

$=3\int_{1}^{b}{x^{-2}}dx$

By the Power Rule for Integrals, $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for all $n\neq-1$ with an arbitrary constant of integration . Therefore:

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

$=3({\frac{x^{-2+1}}{-2+1}})_{1}^{b}$

$=3(\frac{x^{-1}}{-1}})_{1}^{b}$

$=3(-{\frac{1}{x}})_{1}^b$ .

So, $\lim_{b\rightarrow \infty }3[-{\frac{1}{x}}]_{1}^b$

$=\lim_{b\rightarrow \infty }3[-{\frac{1}{b}-(-\frac{1}{1}})]$

$=\lim_{b\rightarrow \infty }3[-{\frac{1}{b}+1]$

As $b\rightarrow \infty$ , $\lim_{b\rightarrow \infty }3[-{\frac{1}{b}+1]=3(0+1)=3$

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

Explanation

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

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

Now, evaluate at 3 and then 1. Subtract the results:

$(27-\frac{9}{2}+3)-(1-\frac{1}{2}+1)=28-4=24$

Evaluate.

$\int \frac{4}{x-1} \ dx$

$F(x) = 4 \ln(x-1)$

$F(x) =\frac{\ln(x-1)}{4}$

$F(x) = \ln(x-1)$

$F(x) = \frac{1}{ \ln(x-1)}$

Answer not listed.

Explanation

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{4}{x-1}$ .

The antiderivative is $F(x) = 4 \ln(x-1)$ .

$\begin{align*}&\text{Evaluate the indefinite integral: }\&\int(\frac{(9e^{(-12x)})}{32})dx\end{align*}$

$-\frac{(3e^{(-12x)})}{128}+C$

$-\frac{(27e^{(-12x)})}{8}$

$-\frac{(27e^{(-12x)})}{8}+C$

$-\frac{(9e^{(-12x)})}{32}$

Explanation

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[e^{ax}]=\frac{e^{ax}}{a}\&\text{Note for this problem, the }a \text{ value is} -12\&\int(\frac{(9e^{(-12x)})}{32})dx=-\frac{(3e^{(-12x)})}{128}+C\&\text{Now you'll note after integration we add a constant, C. This}\&\text{is because we're working with an indefinite integral. Note how}\&\text{if we take the derivative of this function, the constant would}\&\text{go to zero.}\&\end{align*}$

$\begin{align*}&\text{Evaluate the integral: }\&\int_{10}^{15}(\frac{x^{2}}{10})dx\end{align*}$

Explanation

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int x^{a}=\frac{x^{a+1}}{a+1}\&\int_{10}^{15}(\frac{x^{2}}{10})dx=\frac{x^{3}}{30}|_{10}^{15}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{10}^{15}(\frac{x^{2}}{10})dx=\frac{15^{3}}{30}-(\frac{10^{3}}{30})\&\int_{10}^{15}(\frac{x^{2}}{10})dx=79.17\end{align*}$

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