To find the area of the region between $\displaystyle \small f(x)=4-4e^{-2x}$ and the x-axis, integrate the function over the specified interval:

For this particular function use the following rules to integrate.

$\displaystyle \int x^n=\frac{x^{n+1}}{n+1}$ and $\displaystyle \int e^u=\frac{1}{\frac{du}{dx}}e^u$

Therefore we get,

$\displaystyle \small A=\int_0^2(4-4e^{-2x})dx$

$\displaystyle \small A=4x+2e^{-2x}|_0^2$

$\displaystyle \small A=(8+2e^{-4})-(4(0)+2e^{-2(0)})$

$\displaystyle \small A=8+2e^{-4}-2$

$\displaystyle \small \small A=6+2e^{-4}$.

Since the depth of the slab is constant, we know it's volume is given by:

$\displaystyle V= Ah =2A$

The area is, of course given by an integral. To calculate it, we must know where the two functions intersect:

$\displaystyle x+2 = (x^{2}-4)$ 

$\displaystyle x^{2}-x-6 = 0$

Which has roots $\displaystyle x=-2$ and $\displaystyle x=3$. Since we are concerned with area to the right of the y-axis, we discard the negative root. So the area is given by:

 $\displaystyle A=\int_{0}^{3}x+2-(x^{2}-4)dx$

$\displaystyle \int_{0}^{3}-x^{2}+x-6dx$

$\displaystyle [-\frac{1}{3}x^{3}+\frac{1}{2}x^{2}+6x]_{0}^{3}$

=$\displaystyle \frac{27}{2}ft^{2}$

So $\displaystyle V = 2ft(\frac{27}{2}ft^{2})= 27ft^{3}$

The dot product of two vectors can be found by multiplying the first element of one by the first element in the other and adding it to the product of the second elements of each and so on. Using this method the dot product becomes 

$\displaystyle < 9,6,4>\cdot< 3,1,6>=9\cdot3+6\cdot1+4\cdot6$

$\displaystyle =27+6+24=57$

The dot product of two vectors can be found by multiplying the first element of one by the first element in the other and adding it to the product of the second elements of each and so on. Using this method the dot product becomes 

$\displaystyle < 2,1,2>\cdot< 5,5,5>=2\cdot5+1\cdot5+2\cdot5$

$\displaystyle =10+5+10=25$

If we look at the object, we will notice that we have four sides that are added up to give us our total length of string which is $\displaystyle 64$.  This can be expressed as $\displaystyle 2a+2b=64$.  We also know that if we multiply the sides $\displaystyle a$ and $\displaystyle b$, we will find the area, which we are trying to maximize.  This leads us with our answer, $\displaystyle 2a+2b=64$ and $\displaystyle ab=A$. Can you find the length of each side?

To find the area between these two curves, we must start by determining if one function is greater than the other function over the entire interval. Because $\displaystyle \sqrt{x} > x^3$ over the interval $\displaystyle [0,1]$, we can simply subtract the area under $\displaystyle y = x^3$ from the area under $\displaystyle y = \sqrt{x}$. Set up the integration as follows:

$\displaystyle \int_{0}^{1}\sqrt{x} - x^3$

To evaluate this integral, follow the power rule for integrals:

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

Realizing that $\displaystyle \sqrt x = x^{\frac{1}{2}}$, the integral evaluates to:

$\displaystyle (\frac{2\cdot 1^{3/2}}{3} - \frac{1^4}{4}) - (\frac{2\cdot0^{3/2}}{3} - \frac{0^4}{4})$

$\displaystyle \frac{2}{3} - \frac{1}{4}$

$\displaystyle \frac{8}{12} - \frac{3}{12}$

$\displaystyle \frac{5}{12}$

In order to find the area underneath the curve to the x-axis on the interval $\displaystyle [a,b]$ we must solve the integral

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

Because the function is always positive on the interval $\displaystyle [0,15]$ we solve the integral

$\displaystyle \int_{0}^{15} 50 \, dx$.

When taking the integral we apply the inverse power rule which states

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

As such

$\displaystyle \int_{0}^{15} 50 \, dx = 50x|_0^{15}$.

And by the corollary of the first Fundamental Theorem of Calculus

$\displaystyle 50x|_0^{15}=50*15-50*0=750$.

As such the area is

$\displaystyle 750$ square units.

In order to find the area underneath the curve to the x-axis on the interval $\displaystyle [a,b]$ we must solve the integral

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

Because the function is always positive on the interval $\displaystyle [0,2]$ we solve the integral

$\displaystyle \int_{0}^{15} x^4 \, dx$.

When taking the integral we apply the inverse power rule which states

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

As such

$\displaystyle \int_{0}^{2} x^4 \, dx = \frac{x^5}{5}|_0^{2}$.

And by the corollary of the first Fundamental Theorem of Calculus

$\displaystyle \frac{x^5}{5}|_0^{2}=\frac{2^5}{5}-0=\frac{32}{5}$.

As such the area is

$\displaystyle \frac{32}{5}$ square units.

In order to find the area underneath the curve to the x-axis on the interval $\displaystyle [a,b]$ we must solve the integral

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

Because the function is always positive on the interval $\displaystyle [0,1]$ we solve the integral

$\displaystyle \int_{0}^{1} e^x \, dx$.

Because the antiderivative of the exponential function is the exponential function itself, we obtain 

$\displaystyle \int_{0}^{1} e^x \, dx = e^x|_0^1$.

And by the corollary of the first Fundamental Theorem of Calculus

$\displaystyle e^x|_0^1 = e^1-e^0=e-1$.

As such the area is

$\displaystyle e-1$ square units.

Evaluate the following integral to find the area of the region bound by the function and the given limits.

$\displaystyle \int_{0}^{5}x^4-6x^2+8x+6dx$

Begin by recalling the integration rule for polynomials. All we need to do to integrate a term of a polynomial is raise the exponent by 1 and divide by the new number.

So, we go from this

$\displaystyle \int_{0}^{5}x^4-6x^2+8x+6dx$

To this

$\displaystyle \frac{x^5}{5}-\frac{6x^3}{3}+\frac{8x^2}{2}+6x+c|_0^5$

We now have successfully integrate our function, but we still must evaluate it with the given limits. To do so, simply plug in our lower and upper limits and find the difference between them. Note, this is step is simplified by the fact that our lower limit is 0, because plugging zero into our function will just yield "c"

So...

$\displaystyle \frac{5^5}{5}-\frac{6(5)^3}{3}+\frac{8(5)^2}{2}+6(5)+c-c$

$\displaystyle 5^4-2*5^3+4*5^2+30=505$

Making our answer 

$\displaystyle 505 units^2$