Finding Integrals - Math

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$\int_$2^{12}$ x+1{\mathrm{d} x}=?$

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Use the Fundamental Theorem of Calculus: If , then $\int_$a^bf$(x)=g(b)-g(a)$ .

Therefore, we need to find the indefinite integral of our equation first.

To do that, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as since anything to the zero power is one.

$\int x+1{\mathrm{d} $x}=\frac{x^{1+1}$$$}{1+1}+\frac{1x^{0+1}$$}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int x+1{\mathrm{d} $x}=\frac{x^{2}$$$}{2}+\frac{x^{1}$$}{1}+c$

$\int x+1{\mathrm{d} $x}=\frac{1}{2}$x^2$+x+c$

According to FTOC:

$\int_$a^b$ x+1{\mathrm{d} $x}=($\frac{1}{2}$b^2$$+b+c)-($\frac{1}{2}$a^2$+a+c)$

Notice that the 's cancel out.

Plug in our given information and solve.

$\int_$2^{12}$ x+1{\mathrm{d} $x}=($\frac{1}{2}$(12)^2$$+(12))-($\frac{1}{2}$(2)^2$+(2))$

$\int_$2^{12}$ x+1{\mathrm{d} x}=($\frac{1}{2}$144+12)-($\frac{1}{2}$4+2)$

$\int_$2^{12}$ x+1{\mathrm{d} x}=(72+12)-(2+2)$

$\int_$2^{12}$ x+1{\mathrm{d} x}=(84)-(4)$

$\int_$2^{12}$ x+1{\mathrm{d} x}=80$

Evaluate:

$\int_${1}^{e^{100}$} $\frac{1}{x}$dx$

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$\int_${1}^{e^{100}$} $\frac{dx}{x}$ = \ln $e^{100}$ - \ln 1 = 100-0 = 100$

Find $\int_${0}^{\pi}$ $(\cos^{2}$$x+\sin^{2}$x)dx$

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This is most easily solved by recognizing that .

If n is a positive integer, find $\int_${0}^{2\pi}$xsinxdx$ .

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We can find the integral using integration by parts, which is written as follows:

$\int udv=uv-\int vdu$

Let and . We can get the box below:

$\begin{Bmatrix} u=x\ du=dx\ dv=sinxdx\ v=-cosx+c \end{Bmatrix}$

Now we can write:

$\int_${0}^{2\pi}$xsinxdx=\left |-xcosx+cx \right |_${0}^{2\pi }$-\int_${0}^{2\pi}$ (-cosx+c)dx$

$=\left |-xcosx+cx+sinx-cx \right |_${0}^{2\pi }$$

$=(-2\pi +0)-(0+0)$

$=-2\pi$

$\int_${2}^{4}$$x^2${\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our , we can use the reverse power rule to find the indefinite integral or anti-derivative of our function:

$\int $(x^2$){\mathrm{d} $x}=\frac{1}{3}$x^3$+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} $x}=($\frac{1}{3}$b^3$$+c)-($\frac{1}{3}$a^3$+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${2}^{4}$f(x){\mathrm{d} x}=($\frac{1}{3}$\cdot $4^3$)-($\frac{1}{3}$\cdot $8^3$)$

$\int_${2}^{4}$f(x){\mathrm{d} x}=($\frac{64}{3}$)-($\frac{8}{3}$)$

$\int_${2}^{4}$f(x){\mathrm{d} x}=\frac{56}{3}$=18$\frac{2}{3}$$

$\int_${1}^{2}$$\frac{1}{x}${\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

As it turns out, since our $f(x)=\frac{1}{x}$$ , the power rule really doesn't help us. $\frac{1}{x}$ has a special anti derivative: $\ln{x}$ .

$\int $\frac{1}{x}${\mathrm{d} x}=$\ln{x}$+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} x}=($\ln{b}$+c)-($\ln{a}$+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${1}^{2}$f(x){\mathrm{d} x}=(\ln(2))-(\ln(1))$

$\int_${1}^{2}$f(x){\mathrm{d} x}=(\ln(2))-(0)$

$\int_${1}^{2}$f(x){\mathrm{d} x}\approx 0.69$

$\int_${3}^{40}$$e^x${\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

As it turns out, since our , the power rule really doesn't help us. is the only function that is it's OWN anti-derivative. That means we're still going to be working with .

$\int $e^x${\mathrm{d} $x}=e^x$+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} $x}=(e^b$$+c)-(e^a$+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${3}^{40}$f(x){\mathrm{d} $x}=(e^{40}$$)-(e^3$)$

$\int_${3}^{40}$f(x){\mathrm{d} x}=(2.35\cdot $10^{17}$)-(20.09)$

Because is so small in comparison to the value we got for , our answer will end up being $2.35\cdot $10^{17}$$

$\int_$2^6$$(5x^2$+3x+2){\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our , we can use the power rule for all of the terms involved to find our anti-derivative:

$\int $(5x^2$+3x+2){\mathrm{d} $x}=\frac{5}{3}$x^3$$+\frac{3}{2}$x^2$+2x+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} $x}=($\frac{5}{3}$b^3$$+\frac{3}{2}$b^2$$+2b+c)-($\frac{5}{3}$a^3$$+\frac{3}{2}$a^2$+2a+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${2}^{6}$f(x){\mathrm{d} $x}=($\frac{5}{3}$(6)^3$$+\frac{3}{2}$(6)^2$$+2(6))-($\frac{5}{3}$(2)^3$$+\frac{3}{2}$(2)^2$+2(2))$

$\int_${2}^{6}$f(x){\mathrm{d} x}=(360+54+12)-($\frac{40}{3}$+6+4)$

$\int_${2}^{6}$f(x){\mathrm{d} x}=(426)-(23\t$\frac{1}{3}$)$

$\int_${2}^{6}$f(x){\mathrm{d} x}=402\t$\frac{2}{3}$$

$\int_$3^5$$\frac{x+3}{x}${\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\frac{x+3}{x}$$ , we can't use the power rule. We have to break up the quotient into separate parts:

$f(x)=\frac{x}{x}$+\frac{3}{x}$=1+\frac{3}{x}$$ .

The integral of 1 should be no problem, but the other half is a bit more tricky:

$\int$\frac{3}{x}${\mathrm{d} x}$ is really the same as $3\int$\frac{1}{x}${\mathrm{d} x}$ . Since $\int$\frac{1}{x}${\mathrm{d} x}=$\ln{x}$$ , $\int$\frac{3}{x}${\mathrm{d} x}=3$\ln{x}$$ .

Therefore:

$\int$\frac{x+3}{x}${\mathrm{d} x}=x+3$\ln{x}$+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} x}=(b+3$\ln{b}$+c)-(a+3$\ln{a}$+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${3}^{5}$f(x){\mathrm{d} x}=(5+3$\ln{5}$)-(3+3$\ln{3}$)$

$\int_${3}^{5}$f(x){\mathrm{d} x}=(9.83)-(6.30)$

$\int_${3}^{5}$f(x){\mathrm{d} x}\approx 3.53$

$\int_$2^5$$\sqrt{x}$$\text{ }{\mathrm{d}$ x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=$\sqrt{x}$$ , we can use the power rule, if we turn it into an exponent:

This means that:

$\int f(x){\mathrm{d} x}=\int x^$\frac{1}{2}${\mathrm{d} x}=\frac{2}{3}$x^$\frac{3}{2}$+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} x}=($\frac{2}{3}$b^$\frac{3}{2}$+c)-($\frac{2}{3}$a^$\frac{3}{2}$+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${2}^{5}$f(x){\mathrm{d} x}=($\frac{2}{3}$(5)^$\frac{3}{2}$)-($\frac{2}{3}$(2)^$\frac{3}{2}$)$

$\int_${2}^{5}$f(x){\mathrm{d} x}\approx(7.45)-(1.89)$

$\int_${2}^{5}$f(x){\mathrm{d} x}\approx5.56$

$\int_${12}^{15}$\ln(x){\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\ln(x)$ , we can't use the power rule, as it has a special antiderivative:

$\int f(x){\mathrm{d} x}=\int \ln(x){\mathrm{d} x}=x\ln(x)-x+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} x}=(b\ln(b)-b+c)-(a\ln(a)-a+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${12}^{15}$f(x){\mathrm{d} x}=(15\ln(15)-15)-(12\ln(12)-12)$

$\int_${12}^{15}$f(x){\mathrm{d} x}\approx(25.62)-(17.82)$

$\int_${12}^{15}$f(x){\mathrm{d} x}\approx7.8$

$\int_${\pi}^{2\pi}$\sin(x){\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\sin(x)$ , we can't use the power rule, as it has a special antiderivative:

$\int f(x){\mathrm{d} x}=\int \sin(x){\mathrm{d} x}=-\cos(x)+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} x}=(-\cos(b)+c)-(-\cos(a)+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=(-\cos({2\pi}))-(-\cos(\pi))$

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=(-1)-(-(-1))$

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=-2$

$\int_${\pi}^{2\pi}$\cos(x){\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\cos(x)$ , we can't use the power rule, as it has a special antiderivative:

$\int f(x){\mathrm{d} x}=\int \cos(x){\mathrm{d} x}=\sin(x)+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} x}=(\sin(b)+c)-(\sin(a)+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=(\sin({2\pi}))-(\sin(\pi))$

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=(0)-(0)$

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=0$

$\int_${\pi}^{2\pi}$\sin(2x){\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\sin(2x)$ , we can't use the power rule. Instead we end up with:

$\int f(x){\mathrm{d} x}=\int \sin(2x){\mathrm{d} x}=\frac{-\cos(2x)}{2}$+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} x}=($\frac{-\cos(2b)}{2}$+c)-($\frac{-\cos(2a)}{2}$+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=($\frac{-\cos(4\pi)}{2}$)-($\frac{-\cos(2\pi)}{2}$)$

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=($\frac{-1}{2}$)-($\frac{-1}{2}$)$

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=0$

$\int_${\pi}^{2\pi}$\cos(2x){\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_${a}^{b}$f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\cos(2x)$ , we can't use the power rule. Instead we must use u-substituion. If $u = 2x, du = 2dx,and\ dx = $\frac{du}{2}$$ .

$\int f(x){\mathrm{d} x}=\int \cos(2x){\mathrm{d} x}= \int cos(u) \cdot $\frac{1}{2}$dx)= $\frac{sin(u)}{2}$+c=\frac{\sin(2x)}{2}$+c$

Remember to include the for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_${a}^{b}$f(x){\mathrm{d} x}=($\frac{\sin(2b)}{2}$+c)-($\frac{\sin(2a)}{2}$+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=($\frac{\sin(4\pi)}{2}$)-($\frac{\sin(2\pi)}{2}$)$

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=($\frac{0}{2}$)-($\frac{0}{2}$)$

$\int_${\pi}^{2\pi}$f(x){\mathrm{d} x}=0$

$\int_$5^{10}$$\2x^2${\mathrm{d} x}=?$

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Remember the fundamental theorem of calculus!

$\int_$a^bf$(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}- \int f(a){\mathrm{d} x}$

Since our , we can use the reverse power rule to find that the antiderivative is:

Remember to include a for any integral or antiderivative taken!

Now we can plug that equation into our FToC equation:

$\int_$a^bf$(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}- \int f(a){\mathrm{d} x}$

$\int_$a^b$ $2x^2${\mathrm{d} $x}=($\frac{2}{3}$b^3$$+c)-($\frac{2}{3}$a^3$+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_$5^{10}$ $2x^2${\mathrm{d} $x}=($\frac{2}{3}$(10)^3$$)-($\frac{2}{3}$(5)^3$)$

$\int_$5^{10}$ $2x^2${\mathrm{d} x}=(666\t$\frac{2}{3}$)-(88\t$\frac{1}{3}$)$

$\int_$5^{10}$ $2x^2${\mathrm{d} x}=583\t$\frac{1}{3}$$

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Remember the Rundamental Theorem of Calculus: If , then .

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

$\int $x^2$+5x{\mathrm{d} $x}=\frac{x^{2+1}$$$}{2+1}+\frac{5x^{1+1}$$}{1+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int $x^2$+5x{\mathrm{d} $x}=\frac{1}{3}$x^3$$+\frac{5}{2}$x^2$+c$

Now we can plug that back into the problem.

Notice that the 's cancel out. Plug in the values given in the problem:

?

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Remember the fundamental theorem of calculus! If , then .

Since we're given , we need to find the indefinite integral of the equation to get .

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

$\int 5x+8{\mathrm{d} $x}=\frac{5x^{1+1}$$$}{1+1}+\frac{8x^{0+1}$$}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int 5x+8{\mathrm{d} $x}=\frac{5x^{2}$$$}{2}+\frac{8x^{1}$$}{1}+c$

$\int 5x+8{\mathrm{d} $x}=\frac{5}{2}$x^2$+8x +c$

Now we can plug that back in:

Notice that the 's cancel out.

Plug in our given numbers.

$\int_$4^5$ $x^3$+2x+5$ ?

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Remember the fundamental theorem of calculus! If , then .

Since we're given , we need to find the indefinite integral of the equation to get .

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent.

We're going to treat as , as anything to the zero power is one.

For this problem, that would look like:

$\int $x^3$+2x+5{\mathrm{d} $x}=\frac{x^{3+1}$$$}{3+1}+\frac{2x^{1+1}$$$}{1+1}+\frac{5x^{0+1}$$}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int $x^3$+2x+5{\mathrm{d} $x}=\frac{x^{4}$$$}{4}+\frac{2x^{2}$$$}{2}+\frac{5x^{1}$$}{1}+c$

$\int $x^3$+2x+5{\mathrm{d} $x}=\frac{1}{4}$x^4$$+x^2$+5x+c$

Plug that back into the FTOC:

$\int_$a^b$ $x^3$+2x+5{\mathrm{d} $x}=($\frac{1}{4}$b^4$$+b^2$$+5b+c)-($\frac{1}{4}$a^4$$+a^2$+5a+c)$

Notice that the 's cancel out.

Plug in our given values from the problem.

$\int_$4^5$ $x^3$+2x+5{\mathrm{d} $x}=($\frac{1}{4}$(5)^4$$+(5)^2$$+5(5))-($\frac{1}{4}$(4)^4$$+(4)^2$+5(4))$ $\int_$4^5$ $x^3$+2x+5{\mathrm{d} x}=($\frac{1}{4}$625+25+25)-($\frac{1}{4}$256+16+20)$

$\int_$4^5$ $x^3$+2x+5{\mathrm{d} x}=(156.25+25+25)-(64+16+20)$

$\int_$4^5$ $x^3$+2x+5{\mathrm{d} x}=(206.25)-(100)$

$\int_$4^5$ $x^3$+2x+5{\mathrm{d} x}=106.25$

$\int_$4^8$ $6x^2$+8{\mathrm{d} x}=?$

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Use the Fundamental Theorem of Calculus: If , then $\int_$a^bf$(x)=g(b)-g(a)$ .

Therefore, we need to find the indefinite integral of our equation first.

To do that, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as since anything to the zero power is one.

$\int $6x^2$+8{\mathrm{d} $x}=\frac{6x^{2+1}$$$}{2+1}+\frac{8x^{0+1}$$}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int $6x^2$+8{\mathrm{d} $x}=\frac{6x^{3}$$$}{3}+\frac{8x^{1}$$}{1}+c$

$\int $6x^2$+8{\mathrm{d} $x}=2x^3$+8x+c$

That means that $\int_$a^b6x$^2+8{\mathrm{d} $x}=(2b^3$$+8b+c)-(2a^3$+8a+c)$ .

Notice that the 's cancel out.

From here, plug in our numbers.

$\int_$4^86x$^2+8{\mathrm{d} $x}=(2(8)^3$$+8(8))-(2(4)^3$+8(4))$

$\int_$4^86x$^2+8{\mathrm{d} x}=(2512+64)-(264+32)$

$\int_$4^86x$^2+8{\mathrm{d} x}=(1024+64)-(128+32)$

$\int_$4^86x$^2+8{\mathrm{d} x}=(1088)-(160)$

$\int_$4^86x$^2+8{\mathrm{d} x}=928$