Remember the fundamental theorem of calculus!

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

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

\(\displaystyle \int e^x{\mathrm{d} x}=e^x+c\)

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

Because \(\displaystyle e^3\) is so small in comparison to the value we got for \(\displaystyle e^{40}\), our answer will end up being \(\displaystyle 2.35\cdot 10^{17}\)

Remember the fundamental theorem of calculus!

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

Since our \(\displaystyle f(x)=5x^2+3x+2\), we can use the power rule for all of the terms involved to find our anti-derivative:

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

\(\displaystyle \int_{2}^{6}f(x){\mathrm{d} x}=(426)-(23\tfrac{1}{3})\)

\(\displaystyle \int_{2}^{6}f(x){\mathrm{d} x}=402\tfrac{2}{3}\)

Remember the fundamental theorem of calculus!

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

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

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

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

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

Therefore:

\(\displaystyle \int\frac{x+3}{x}{\mathrm{d} x}=x+3\ln{x}+c\)

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Remember the fundamental theorem of calculus!

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

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

\(\displaystyle f(x)=\sqrt{x}=x^{\frac{1}{2}}\)

This means that:

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Remember the fundamental theorem of calculus!

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

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

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Remember the fundamental theorem of calculus!

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

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

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Remember the fundamental theorem of calculus!

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

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

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Remember the fundamental theorem of calculus!

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

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

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Remember the fundamental theorem of calculus!

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

Since our \(\displaystyle f(x)=2x^2\), we can use the reverse power rule to find that the antiderivative is:

 \(\displaystyle \int2x^2{\mathrm{d} x}=\frac{2}{3}x^3+c\)

Remember to include a \(\displaystyle +c\) for any integral or antiderivative taken!

Now we can plug that equation into our FToC equation:

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

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

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

\(\displaystyle \int_5^{10} 2x^2{\mathrm{d} x}=(\frac{2}{3}(10)^3)-(\frac{2}{3}(5)^3)\)

\(\displaystyle \int_5^{10} 2x^2{\mathrm{d} x}=(666\tfrac{2}{3})-(88\tfrac{1}{3})\)

\(\displaystyle \int_5^{10} 2x^2{\mathrm{d} x}=583\tfrac{1}{3}\)

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

\(\displaystyle \int udv=uv-\int vdu\)

Let \(\displaystyle u=x\) and \(\displaystyle dv=sinxdx\). We can get the box below:

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

Now we can write:

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

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

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

\(\displaystyle =-2\pi\)