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

This is most easily solved by recognizing that $\displaystyle \sin^{2}x+\cos^{2}x=1$.  

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

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

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

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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 $\displaystyle f(x)=x^2$, we can use the reverse power rule to find the indefinite integral or anti-derivative of our function:

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

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

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Remember the fundamental theorem of calculus!

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

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

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

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

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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:

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

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

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