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.

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

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

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:

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

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

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

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

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 $\displaystyle 5$ as $\displaystyle 5x^0$, as anything to the zero power is one.

For this problem, that would look like:

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

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

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

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

Set $\displaystyle u = \sin x$. Then 

$\displaystyle \frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \sin x= \cos x$.

and 

$\displaystyle \cos x \; dx = du$

The integral becomes:

$\displaystyle \dpi{120} \int\cos x \sqrt{\sin x }\; \; dx$

$\displaystyle \dpi{120} =\int \sqrt{u }\; \; du$

$\displaystyle = \int u ^{\frac{1}{2}} \; du$

$\displaystyle \dpi{120} \dpi{150} = \frac{u^{\frac{1}{2}+1}}{{\frac{1}{2}+1}} + C$

$\displaystyle \dpi{150} = \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C$

$\displaystyle \dpi{120} =\frac{ 2u \sqrt{u}} {3} +C$

Substitute back:

$\displaystyle \dpi{120} =\frac{ 2 \sin x \cdot \sqrt{\sin x }} {3} +C$

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:

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

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

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

To find the indefinite integral of our expression, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the $\displaystyle x$ by one and then divide by that new exponent.

First we need to realize that $\displaystyle 2x=2x^1$. From there we can solve:

$\displaystyle \int2x{\mathrm{d} x}=\frac{2x^{1+1}}{1+1}+c$

When taking an integral, be sure to include a $\displaystyle +c$ at the end of everything. $\displaystyle c$ stands for "constant". Since taking the derivative of a constant whole number will always equal $\displaystyle 0$, we include the $\displaystyle +c$ to anticipate the possiblity of the equation actually being $\displaystyle x^2+5$ or $\displaystyle x^2-100$ instead of just  $\displaystyle x^2$.

$\displaystyle \int2x{\mathrm{d} x}=\frac{2x^{2}}{2}+c$

$\displaystyle \int2x{\mathrm{d} x}=x^2+c$