Set up your integral using the given bounds, then solve.

Remember the rules of trigonometric functions,

$\displaystyle \int sin(x)=-cos(x)$.

Therefore our equation becomes,

$\displaystyle \\ \int_{-\pi}^\pi sin(x) dx=-cos(x)]_{-\pi}^\pi\\ \\=-cos(\pi)-(-cos(-\pi))\\ \\=-(-1)-(-(-1))\\ \\=1-(1)=0$.

Set up your integral using the given bounds, then solve by using the power rule

$\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}$: 

$\displaystyle \\ \int_{0}^4 \sqrt x dx=\int_0^4x^{\frac{1}{2}} dx \\ \\=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\left|_0^4=\frac{2x^{\frac{3}{2}}}{3}\left|_0^4\\ \\=\frac{2(4)^{\frac{3}{2}}}{3}=\frac{2\cdot8}{3}\\ \\=\frac{16}{3}$.

First, graph the two functions in order to identify the boundaries of the region. You will find that they are $\displaystyle [-2,2]$.

Therefore, when you set up your integral, it will be from $\displaystyle -2$ to $\displaystyle 2$.

Then solve the integral by using the power rule

$\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}$: 

$\displaystyle \\ \int_{-2}^2-x^2+4dx=-\int_{-2}^2x^2dx+\int_{-2}^24dx\\ \\=-\frac{x^3}{3}|_{-2}^2+4x|_{-2}^2\\ \\=-\left[\frac{(2)^3}{3}-\frac{(-2)^3}{3}\right]+\left[4(2)-4(-2)\right]\\ \\=-\left[\frac{8}{3}-\frac{-8}{3}\right]+[8+8]\\ \\=-\left[\frac{16}{3}\right]+16\\ \\=\frac{32}{3}$

Set up your integral using the given bounds, then solve by using the power rule

$\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}$: 

$\displaystyle \\ \int_0^3x+2dx=\int_0^3xdx+\int_0^32dx\\ \\=\frac{x^2}{2}|_0^3+2x|_0^3\\ \\=\left(\frac{(3)^2}{2}-0\right)+(2(3)-0)\\ \\=\frac{9}{2}+6\\ \\=\frac{21}{2}$

Set up your integral using the given bounds, then solve by using the power rule

$\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}$: 

$\displaystyle \\ \int_0^6 2x+3 dx=\int_0^6 2x dx+\int_0^6 3 dx\\ \\=\frac{2x^2}{2}|_0^6+3x|_0^6\\ \\=(6)^2-(0)^2+[3(6)^2-3 (0)2]\\ \\=36+18\\ \\=54$.

For this problem, we must first find the upper bound of the region, the x-value where the two curves intersect:

$\displaystyle f(x_u)=g(x_u)$

$\displaystyle 10-x_u^3=\frac{1}{2}x_u^2$

This equation holds for the value of $\displaystyle x_u=2$

Therefore the lower and upper bounds are $\displaystyle x_l=0,x_u=2$. The lower bound is known since we're told the region is bounded by the y-axis.

The area of the region is thus:

$\displaystyle A=\int_{0}^{2}10-x^3-\frac{1}{2}x^2$

$\displaystyle A=10x-\frac{x^4}{4}-\frac{x^3}{6}|_0^2$

$\displaystyle A=(20-4-\frac{4}{3})-(0-0-0)$

$\displaystyle A=\frac{44}{3}$

We slice the region into $\displaystyle n$ thin vertical strips of thickness $\displaystyle \Delta x$ and height $\displaystyle \sin(x_{k})$ and then sum up all $\displaystyle n$ strips, each of area $\displaystyle \sin(x_{k})\Delta x$.

This gives us an approximate expression for the area:

 $\displaystyle Area\approx \sum_{k=1}^{n}\sin(x_{k})\Delta x$ where $\displaystyle x_{k}=0+k\Delta x$ and  $\displaystyle \Delta x= \frac{\pi/3}{n}$.

We take the limit as the number of slices approaches infinity over this interval and we get the definite integral:

$\displaystyle \\ Area=\lim_{n\to\infty}\sum_{k=1}^{n}\sin{x_{k}}\Delta x\\ \\=\int_{0}^{\pi/3}\sin(x)dx\\ \\=-\cos(\pi/3)-[-\cos(0)])\\ \\=-\frac{1}{2}+1=\frac{1}{2}$

To find the area under the curve

$\displaystyle f(x)=20-3x^2$

Integrate it from the specified bounds:

$\displaystyle A=\int_{1}^{2}(20-3x^2)dx=20x-x^3|_1^2$

$\displaystyle A=(40-8)-(20-1)=32-19$

$\displaystyle A=13$

The first step is determine the lower and upper x-values that define the area. There is a lower bound of zero that marks the transition for f(x) to move into negative y-values; however, g(x) is well into the negative at this point, so it'll be necessary to find a lower bound where it first begins to become negative. This will occur for a value of five:

$\displaystyle g(5)=2(5)-10=0$

This allows the creation of an initial integral:

$\displaystyle A_1=\int_{0}^{5}\frac{1}{3}xdx$

$\displaystyle A_1=\frac{x^2}{6}|_0^5$

$\displaystyle A_1=\frac{25}{6}-0=\frac{25}{6}$

Another upper bound can be found by determining the point where the two functions intersect:

$\displaystyle \frac{1}{3}x=2x-10$

$\displaystyle \frac{5}{3}x=10$

$\displaystyle x=6$

Now, integrate the difference of these functions over these final bounds:

$\displaystyle A_2=\int_{5}^{6}\frac{1}{3}x-(2x-10)dx$

$\displaystyle A_2=\frac{x^2}{6}-x^2+10x|_5^6$

$\displaystyle A_2=(\frac{36}{6}-36+60)-(\frac{25}{6}-25+50)$

$\displaystyle A_2=\frac{11}{6}-11+10=\frac{5}{6}$

The full area is now the sum of these two:

$\displaystyle A=A_1+A_2$

$\displaystyle A=\frac{25}{6}+\frac{5}{6}=\frac{30}{6}$

$\displaystyle A=5$

In order to find the area under $\displaystyle f(x)$ on the interval of $\displaystyle x=0$ to $\displaystyle x=4$, you must evaluate the definite integral

 $\displaystyle \int_{}0}^{4}f(x)dx{}$

$\displaystyle =\int_{}0}^{4}e^{2x}dx$

First, antidifferentiate the function.

$\displaystyle =-\frac{1}{2}e^{2x}\mid_{0}^{4}{}$

Then, substitute values for $\displaystyle x$.

$\displaystyle =\frac{1}{2}e^{2*4}-\frac{1}{2}e^{2*0}$

Finally, evaluate in terms of $\displaystyle e$

$\displaystyle =\frac{1}{2}e^{2*4}-\frac{1}{2}*(1)$

$\displaystyle =\frac{1}{2}e^{8}-\frac{1}{2}$