Finding the area under the curve can be understood as taking the integral of the equation and it can be rewritten as the following: 

\(\displaystyle \int_{-2}^{3} |4-x^2|dx\)

There are 2 things to note.

1. This function is positive only from \(\displaystyle x=-2\) to \(\displaystyle x=2\). From \(\displaystyle x=2\) to \(\displaystyle 3\), this function is negative.

2. This function is not differentiable at \(\displaystyle x=2\) because of the absolute value.

Therefore, this function needs to be broken up into 2 pieces.

\(\displaystyle \int_{-2}^{2} |4-x^2|dx + \int_{2}^{3} x^2-4dx\)

To solve:

1. Find the indefinite integral of the function.

2. Plug in the upper and lower limit values and take the difference of the two values.

1. Using the power rule which states,

\(\displaystyle \int x^n =\frac{x^{n+1}}{n+1}\) to each term we find,

\(\displaystyle \int 4 - x^2dx = 4x - \frac{x^3}{3} +constant\)

\(\displaystyle \int x^2-4dx =\frac{x^3}{3} -4x +constant\).

2. Plug in the respective upper and lower limits for x and then take the difference.

\(\displaystyle 4x - \frac{x^3}{3}|_{-2}^{2} = 8-\frac{8}{3}-\left(-8 + \frac{8}{3}\right) = 16 - \frac{16}{3} = \frac{32}{3}\)

\(\displaystyle \frac{x^3}{3} - 4x|_{2}^3 = \frac{27}{3} - 12 -\left(\frac{8}{3}-8\right) = 9 - 12 + 8 - \frac{8}{3} = 5 - \frac{8}{3}\)

Now take the sum of the integrals of the two pieces.

\(\displaystyle \frac{32}{3}+ 5 - \frac{8}{3} = 5 + \frac{24}{3} = 5 + 8 = 13\)

Find the area under \(\displaystyle y=\sin{x}\) where \(\displaystyle 0< x< \pi/2\).

Take the integral of \(\displaystyle f(x)=\sin{x}\):

\(\displaystyle F(x)=-\cos{x}\)

Next, evaluate it on the interval:

\(\displaystyle F(\pi/2)-F(0)\)

\(\displaystyle (-\cos(\pi/2))-(-\cos(0))\)

\(\displaystyle (-0)-(-1)\)

\(\displaystyle 1\)

To find this area, you can either take the integral of \(\displaystyle ln(x)\) from 1 to \(\displaystyle e\), or you can redefine the integral in terms of \(\displaystyle y\) and solve that way, as the integral of the natural log is more difficult to find.

To do this, rewrite \(\displaystyle y=ln(x)\) as \(\displaystyle e^y=x\) and then redefine the bounds of the integral, so instead of integrating on \(\displaystyle (1,e)\) you can integrate on \(\displaystyle (0,1)\).

The final integral is 

\(\displaystyle \int_0^1 e^y dy= e-1\).

This however, is the area between the curve and the y-axis. To find the remaining area, subtract this from the area from \(\displaystyle e\) to the \(\displaystyle y\)-axis on \(\displaystyle (0,1)\), which has an area of \(\displaystyle e\).

Thus, the final expression resulting in the answer is,

 \(\displaystyle e-(e-1)=1\) .

To find the area under the curve, we need to perform a definite integral. Essentially, this integral will be summing up all the infinitesimally small rectangles that make up the region. The entire region is in the first quadrant, so we don't have to worry about splitting our region up. 

When we take the integral we will need to use,

\(\displaystyle \int x^n=\frac{x^{n+1}}{n+1}\) then plug in the upper and lower bounds into the function and take the difference.

Therefore,

\(\displaystyle \small \int_{0}^{5}(x^2)dx=(5)^3/3-0^3/3=125/3\)

To find the area under the curve, evaluate 

\(\displaystyle \int_{0}^{2}f(x)dx\).

\(\displaystyle \int_{0}^{2} 2cos\left(\frac{\pi x}{4}\right)dx\)

Remember the integral of \(\displaystyle cos(x)\) is \(\displaystyle sin(u)\cdot \frac{1}{\frac{du}{dx}}\).

Integrating yields:

\(\displaystyle = \frac{8}{\pi}sin\left(\frac{\pi x}{4}\right)|_{0}^{2}\)

Now, evaluate:

\(\displaystyle = \frac{8}{\pi}sin\left(\frac{\pi\cdot 2 }{4}\right) - \frac{8}{\pi}sin(0) = \frac{8}{\pi}sin\left(\frac{\pi }{2}\right)-0\)

\(\displaystyle =\frac{8}{\pi}\)

 If \(\displaystyle f(x)\geq g(x)\) on the interval \(\displaystyle a\leq x \leq b\), then the area bounded by the two functions is given by 

\(\displaystyle \int_a^b (f(x)-g(x))dx\).

To find the area bounded by \(\displaystyle x+1\) and \(\displaystyle x^2+1\), we must first find when they intersect.  

Setting the functions equal \(\displaystyle x+1=x^2+1\), we get \(\displaystyle 0=x^2-x=x(x-1)\).

Therefore the two functions intersect at \(\displaystyle x=0\) and \(\displaystyle x=1\) and we wish to find the area of the region bounded by \(\displaystyle f(x)\) and \(\displaystyle g(x)\) between these two points. By plotting the two functions, we see that on this interval, \(\displaystyle x+1\geq x^2+1\).

Therefore, the area of the region bounded by these functions is given by

\(\displaystyle \int_0^1 \left((x+1)-(x^2+1) \right )dx =\int_0^1 \left(-x^2+x \right )dx\)

\(\displaystyle =\left[-\frac{x^3}{3}+\frac{x^2}{2} \right ]_0^1=\left(-\frac{1^3}{3}+\frac{1^2}{2} \right )-\left( -\frac{0^3}{3}+\frac{0^2}{2}\right )\)

\(\displaystyle =-\frac{1}{3}+\frac{1}{2}=\frac{1}{6}\).

The area for the region bounded by the graph of \(\displaystyle f(x)\) and the x-axis on the interval \(\displaystyle [a,b]\) is given as

\(\displaystyle \int_{a}^{b}\left | f(x)\right | dx\).

Because \(\displaystyle f(x)>0\) on the interval \(\displaystyle [0,16]\) the area is

 \(\displaystyle \int_{0}^{16}f(x)\,dx=\int_{0}^{16}x^\frac{1}{2}\,dx\).

When taking the integral, we will use the inverse power rule which states,

\(\displaystyle \int x^n=\frac{x^{n+1}}{n+1}\) .

As such,

\(\displaystyle \int_{0}^{16}x^\frac{1}{2}\,dx=\frac{x^{3/2}}{3/2}\,\left.\begin{matrix} \, \end{matrix}\right|_{0}^{16}=\frac{2}{3}x^\frac{3}{2}\left.\begin{matrix} \, \end{matrix}\right|_{0}^{16}\)

And by the corollary of the First Fundamental Theorem of Calculus,

\(\displaystyle \frac{2}{3}x^\frac{3}{2}\left.\begin{matrix} \, \end{matrix}\right|_{0}^{16}=\left[\frac{2}{3}(16)^{\frac{3}{2}}\right]-\left[\frac{2}{3}(0)^{\frac{3}{2}}\right]\).

Hence the area is

\(\displaystyle \frac{128}{3}\) square units.

The area for the region bounded by the graph of \(\displaystyle f(x)\) and the x-axis on the interval \(\displaystyle [a,b]\) is given as

\(\displaystyle \int_{a}^{b}\left | f(x)\right | dx\).

As such,

\(\displaystyle A=\int_{-2}^{0}-x\, dx \, \, \, + \int_{0}^{2}x \, dx\).

And because the two regions are symmetric about the y-axis,

\(\displaystyle A=2\int_{0}^{2}x \, dx\).

When taking the integral, we will use the inverse power rule which states,

 \(\displaystyle \int x^n=\frac{x^{n+1}}{n+1}\).

As such,

\(\displaystyle A=2\int_{0}^{2}x \, dx= 2\left(\frac{1}{2}x^2\right)\left.\begin{matrix} \, \end{matrix}\right|_{0}^{2}\).

And by the corollary of the First Fundamental Theorem of Calculus,

\(\displaystyle 2\left(\frac{1}{2}x^2\right)\left.\begin{matrix} \, \end{matrix}\right|_{0}^{2}=2\left(\left[\frac{1}{2}(2)^{2}\right]-\left[\frac{1}{2}(0)^2\right]\right)\).

Hence the area is \(\displaystyle 4\) units squared.

The area under the curve of any function is given by the integral of the function. Note that the function evaluated from x=0 to x=1 is always positive, so the integration over the entire interval can be done using one integral. The integral of the function is given by

\(\displaystyle \int_{0}^{1}(3x^4+9x^3+17x+1)dx\) 

and is equal to 

\(\displaystyle \frac{3x^5}{5}+\frac{9x^4}{4}+\frac{17x^2}{2}+x+C\).

The integration comes from the rules 

\(\displaystyle \int u^ndu=\frac{1}{n+1}u^{n+1}+C, n\neq-1\) and \(\displaystyle \int du=u+C\).

Evauating the integral between the two points is done by plugging in the upper limit of integration (x=1) into the function and then subtracting by the function when the lower limit of integration (x=0) is plugged in. When evaluated from the point x=0 to x=1, the integral is equal to 

\(\displaystyle \left(\frac{3}{5}+\frac{9}{4}+\frac{17}{2}+1\right)-(0)=\frac{247}{20}\).

The dot product of \(\displaystyle a=\left \langle 4,-1,3\right \rangle\) and \(\displaystyle b=\left \langle 2,7,5\right \rangle\) is the sum of the products of its individual corresponding components, or

\(\displaystyle (4\times 2)+(-1\times 7)+(3\times 5)=8-7+15=1+15=16\).