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:

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

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

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

The integral of $\displaystyle \frac{1}{x}$ is $\displaystyle \ln \left | x \right |+C$.  The constant 3 is simply multiplied by the integral.  

To integrate $\displaystyle \cos(x)\sin(x)$, we need to get the two equations in terms of each other. We are going to use "u-substitution" to create a new variable, $\displaystyle u$, which will equal $\displaystyle \cos(x)$.

Now, if $\displaystyle u=\cos(x)$, then 

$\displaystyle \frac{\mathrm{d}u }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-sin(x)$ 

Multiply both sides by $\displaystyle \mathrm d {x}$ to get the more familiar: 

$\displaystyle {\mathrm{d} u}=-\sin(x)\mathrm{d}x$

Note that our $\displaystyle \mathrm d{u}=-\sin(x)$, and our original equation was asking for a positive $\displaystyle \sin(x)$.

That means if we want $\displaystyle \int\cos(x)\sin(x)$ in terms of $\displaystyle u$, it looks like this:

$\displaystyle \int u\cdot (-\mathrm d{u}})$

Bring the negative sign to the outside:

$\displaystyle -\int u\mathrm d{u}$.

We can use the power rule to find the integral of $\displaystyle u$:

$\displaystyle -(\frac{1}{2}u^2+c)$

Since we said that $\displaystyle u=\cos(x)$, we can plug that back into the equation to get our answer:

$\displaystyle -(\frac{1}{2}\cos^2(x)+c)$

In this case we have a rational function as $\displaystyle \frac{N(x)}{D(x)}$, where

$\displaystyle N(x)=1$

and

 $\displaystyle D(x)=\frac{1}{x^2-16}$

$\displaystyle D(x)$ can be written as a product of linear factors:

$\displaystyle \frac{1}{x^2-16}=\frac{1}{(x-4)(x+4)}\equiv \frac{A}{x-4}+\frac{B}{x+4}$

It is assumed that A and B are certain constants to be evaluated. Denominators can be cleared by multiplying both sides by (x - 4)(x + 4). So we get:

$\displaystyle 1=A(x+4)+B(x-4)$

First we substitute x = -4 into the produced equation:

$\displaystyle 1=-8B\Rightarrow B=-\frac{1}{8}$

Then we substitute x = 4 into the equation:

$\displaystyle 1=8A\Rightarrow A=\frac{1}{8}$

Thus:

$\displaystyle \frac{1}{(x-4)(x+4)}=\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4}$

Hence:

$\displaystyle \int \frac{dx}{x^2-16}=\int(\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4})dx$

$\displaystyle =\frac{1}{8}ln\left | x-4 \right |-\frac{1}{8}ln\left |x+4 \right |+c$

$\displaystyle =\frac{1}{8}(ln\left | x-4 \right |-ln\left |x+4 \right |)+c$

$\displaystyle =\frac{1}{8}ln\left | \frac{x-4}{x+4}\right |+c$

$\displaystyle \log_{10} x = \frac{\ln x}{\ln 10}$, so this can be rewritten as 

$\displaystyle \dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}\; \; dx$

$\displaystyle \dpi{120}=\int\frac{\sqrt{\ln x}}{x \sqrt{\ln 10} }\; \; dx$

$\displaystyle \dpi{120}= \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

Set $\displaystyle u = \ln x$. Then

$\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and 

$\displaystyle du = \frac{dx}{x}$

Substitute:

$\displaystyle \dpi{120} \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

$\displaystyle \dpi{120} =\frac{1}{\sqrt{\ln 10}} \int\sqrt{u} \; du$

$\displaystyle \dpi{120} =\frac{1}{\sqrt{\ln 10}} \int u ^{\frac{1}{2}} \; du$

$\displaystyle \dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C \right )$

$\displaystyle \dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2u \sqrt{u}} {3} + C \right )$

The outer factor can be absorbed into the constant, and we can substitute back:

$\displaystyle \dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2 \ln x \cdot \sqrt{ \ln x}} {3} \right )+ C$

$\displaystyle = \frac{ 2 \sqrt{ \ln x} \cdot \ln x} {3 \cdot \sqrt{\ln 10}} + C$

$\displaystyle = \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

Set $\displaystyle u = \ln x$. Then

$\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and 

$\displaystyle du = \frac{dx}{x}$

Also, since $\displaystyle u = \ln x$, the limits of integration change to $\displaystyle \ln e^{3} = 3$ and $\displaystyle \ln 1= 0$.

Substitute:

$\displaystyle \dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}\; \; dx$

$\displaystyle \dpi{120} = \int_{0}^{3}\sqrt{u} \; du$

$\displaystyle = \int_{0}^{3} u ^{\frac{1}{2}} \; du$

$\displaystyle \dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \\ & \end{matrix}\right|$ $\displaystyle \begin{matrix} 3\\ \\ 0 \end{matrix}$

$\displaystyle \dpi{150} = \left.\begin{matrix} \\ \frac{ 2u \sqrt{u}} {3} & \\ & \end{matrix}\right|$ $\displaystyle \begin{matrix} 3\\ \\ 0 \end{matrix}$

$\displaystyle =\frac{ 2\cdot 3\cdot \sqrt{3}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} =2\sqrt{3}$

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 $\displaystyle f(x)=x^2$, 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 $\displaystyle 2.35\cdot 10^{17}$

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:

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

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

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

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