The area of the region between a function and the x-axis on the interval \(\displaystyle [a,b]\)  is given as the definite integral

\(\displaystyle \int_a^b |f(x)| \,dx\)

Because  \(\displaystyle f(x)>0\)  on the interval \(\displaystyle [3,7]\) the area formula becomes

\(\displaystyle \int_3^7 x^2-6x+9\,dx\)

In order to solve the integral we use the inverse power rule which says

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

And because integrating is a linear operation we apply the power rule term-by-term to get

\(\displaystyle \int_3^7 x^2-6x+9\,dx=(\frac{x^3}{3}-6*\frac{x^2}{2}+9x|_3^7\)

\(\displaystyle =(\frac{x^3}{3}-3x^2+9x|_3^7\)

And by the corollary of the Fundamental Theorem of Calculus

\(\displaystyle \frac{x^3}{3}-3x^2+9x|_3^7=[\frac{7^3}{3}-3(7)^2+9(7)]-[\frac{3^3}{3}-3(3)^2+9(3)]\)

\(\displaystyle =[\frac{343}{3}-147+63]-[9-27+27]\)

\(\displaystyle =\frac{64}{3}\)

Find the area of the region bound by the y-axis, h(t), and the lines \(\displaystyle t=5\), \(\displaystyle t=6\)

\(\displaystyle h(t)=5e^t+18t^2-13\)

First, we want to set up our integral:

\(\displaystyle \int_{5}^{6}5e^t+18t^2-13dt\)

Next, recall that \(\displaystyle e^t\) integrates to \(\displaystyle e^t\), and that to integrate a polynomial, we simply take each term, increase its exponent by 1, and divide by the new number. Doing so yields the following:

\(\displaystyle H(t)=\int_{5}^{6}h(t)dt=5e^t+6t^3-13t+c\mid_5^6\)

Now, to find the area, we need to find the H(6)-H(5)

\(\displaystyle (5e^6+6(6)^3-13(6)+c)-(5e^5+6(5)^3-13(5)+c)\)

Simplify to get:

\(\displaystyle (2017.144+1296-78+c)-(742.066+750-65+c)=3235.144-1427.066\)

\(\displaystyle 3235.144-1427.066=1808.078\)

So, we can round our answer to get:

\(\displaystyle 1808\)

Set this up as a definite integral to find volume. In order to do this, we need to know which function is above the other. In this case,  \(\displaystyle 1\geq \frac{1}{x}\) for all \(\displaystyle x:(1,2)\). 

To solve for area \(\displaystyle A\): 

\(\displaystyle A=\int_{1}^{2} [1-\frac{1}{x} ]dx\)

Notice that \(\displaystyle \int \frac{1}{x}=ln(x)\). 

Therefore:

\(\displaystyle \int 1-\frac{1}{x} dx=x-ln(x)=F(x)\) 

By the fundamental theorem of calculus:

\(\displaystyle A=\int_{1}^{2} [1-\frac{1}{x} ]dx=F(2)-F(1)=(2-ln(2))-(1-ln(1))=1-ln(2)\)

We can find the interval that we are supposed to integrate over by setting the two equations equal to each other.

\(\displaystyle 4x=(x-3)^2\) 

Manipulating the equation you get 

\(\displaystyle 0= x^2-10x+9\).  

Factoring this gives you, 

\(\displaystyle 0=(x-9)(x-1)\).  

Therefore we integrate between 1 and 9.  

Next we see that if we were to graph both equations, \(\displaystyle 4x\) is above \(\displaystyle (x-3)^2\) and we know that we need to have the upper function minus the lower so our integral will be the following: \(\displaystyle \\ \int_1^9 (4x-(x-3)^2)dx\\\\= \int_1^9(-x^2+10x-9)dx\\\\= (-\frac{1}{3}x^3+5x^2-9x)|_1^9\\\\=85.33\)

There are two different ways to solve this problem.  You can integrate both pieces of the piecewise function separately and add them together to get the area or you can look at the problem in a more geometric way.  

First we graph the equation:

.  

We can actually make this into two right triangles

 .  

The area of a triangle is known as

\(\displaystyle A_{triangle}=\frac{1}{2}bh\).  

The red triangle has a height of \(\displaystyle f(-2)=6\) and a base of length two.  

The blue triangle has a height of \(\displaystyle f(1)=4\) and a base of one.  

Therefore the area of this function on the interval \(\displaystyle [-2, 1]\) is 

\(\displaystyle \frac{1}{2}\cdot 2 \cdot 6+\frac{1}{2}\cdot 1 \cdot 4=6+2=8\).

The first step is to write the first equation as a function of x.  You will note however that this is the equation for a circle and is thus not a function of x.  This ends up not being a problem however since \(\displaystyle y=x^2\) is only located in quadrant 1 and 2 and thus we only need the piece of the circle in quadrant 1 and 2 which happens to be a function, namely \(\displaystyle y=\sqrt{6-x^2}\).  

The next step is to find the points in which these two functions intersect by setting them equal to each other.  

This yields the points \(\displaystyle (-\sqrt 2, 2) \text{ and } (\sqrt 2, 2)\). All that is left now is to plug them into the correct formula: 

\(\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (\sqrt{6-x^2}-x^2)dx\).  

Yet since it is symmetric, we can rewrite it as 

\(\displaystyle 2\int_{0}^{\sqrt{2}} (\sqrt{6-x^2}-x^2)dx\).