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

\(\displaystyle f(x)=\frac{16}{x}\)

To find area, set up an integral:

\(\displaystyle \int_{5}^{6}\frac{16}{x}dx\)

Recall that:

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

So we get:

\(\displaystyle \int_{5}^{6}\frac{16}{x}dx=16ln(x)^6_5\)

\(\displaystyle 16ln(6)-16ln(5)=2.917\)

So our answer is:

\(\displaystyle 2.917\)

To solve this problem we must be able to find the point on the graph of the equation that is the maximum.  

In order to achieve this we must take the derivative of the equation, set it equal to zero and solve for \(\displaystyle x\).  

Taking the derivative of the equation using the power rule 

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}}\),

we find that the equation becomes 

\(\displaystyle y'=-26x+770718\).  

Setting the equation equal to 0 and solving for \(\displaystyle x\), we find that in order to maximize profits, the company should make 29,643 printers a year.

This can be integrated through integration by parts. 

\(\displaystyle u=2x\rightarrow u'=2dx\)

\(\displaystyle v'=cos(4x)\rightarrow v=\frac{sin(4x)}{4}\)

\(\displaystyle \\\int 2xcos(4x)dx\\ \\=uv-\int u'v\\ \\=\frac{xsin(4x)}{2}-\int \frac{sin(4x)dx}{2}\)

\(\displaystyle =\frac{xsin(4x)}{2}+\frac{cos(4x)}{8}+C\)

Let \(\displaystyle u=x^2\rightarrow du=2xdx\)

Therefore by u substitution we are able to integrate thus finding the area under the curve. Once you have integrated remember to plug your original variables back into the equation:

\(\displaystyle \\ \int 2xcos(x^2)dx\\ \\=\int cos(u)du\\ \\=sin(u)+C\\ \\=sin(x^2)+C\)

 Looking at the graph, we see that the graphs intersect at the point \(\displaystyle (1,1)\) with \(\displaystyle y=\sqrt{x}\) on top.

Therefore, our integral to find the area by partitioning the x-axis looks like 

\(\displaystyle \int_{0}^{1} [ \sqrt{x}-x^3]dx = \frac{5}{12}\).

First we find that the equation crosses the y-axis at y=-1, 1.

Partitioning the y-axis, we get the area is equal to 

\(\displaystyle \int_{-1}^{1}(1-y^2) dy=\frac{4}{3}\).

To find the area between the 2 trigonometric waves, we have to look at two different regions between 0 and \(\displaystyle \frac{\pi}{2}\).

Cosine is above sine from 0 to \(\displaystyle \frac{\pi}{4}\), and sine is above cosine from \(\displaystyle \frac{\pi}{4}\) to \(\displaystyle \frac{\pi}{2}\):

Written as an integral, this area can be represented as follows:

\(\displaystyle \int_{0}^{\frac{\pi}{4}}cosx - sinx dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}sinx - cosx dx\)

Integrating yields:

\(\displaystyle sinx - (-cosx)\) from 0 to \(\displaystyle \frac{\pi}{4}\) and \(\displaystyle -cosx - sinx\) from \(\displaystyle \frac{\pi}{4}\) to \(\displaystyle \frac{\pi}{2}\)

Evaluate:

\(\displaystyle sin\left(\frac{\pi}{4}\right)+cos\left(\frac{\pi}{4}\right)-[sin(0)+cos(0)] + \left[-cos\left(\frac{\pi}{2}\right)-sin\left(\frac{\pi}{2}\right)\right]-\left[-cos\left(\frac{\pi}{4}\right)-sin\left(\frac{\pi}{4}\right)\right]\)

\(\displaystyle \left[\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\right]-\left[(0+1) \right ] + [0-1]\left(-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\right)\)

\(\displaystyle \frac{2}{\sqrt2}-1-1+\frac{2}{\sqrt2}\)

\(\displaystyle \frac{4}{\sqrt2}-2\) rationalizing the denominator:

\(\displaystyle \frac{4\sqrt2}{2}-2=2\sqrt2 -2\)

The area between the two curves goes between the two points of intersection:

To figure out the left and right bounds of this region, solve the system of equations \(\displaystyle y=f(x), y=g(x)\) by setting the two equations equal to each other and solving for x:

\(\displaystyle -x^2 +5 = 3x^2 + 1\) add \(\displaystyle x^2\) to both sides

\(\displaystyle 5=4x^2 + 1\) subtract 1 from both sides

\(\displaystyle 4=4x^2\) divide by 4

\(\displaystyle 1=x^2\)

The solutions are -1 and 1, so we're finding the area for \(\displaystyle -1\leq x \leq1\).

For this region, the function \(\displaystyle -x^2 + 5\) is above \(\displaystyle 3x^2 + 1\), so our integral looks like this:

\(\displaystyle \int_{-1}^{1} -x^2+5- (3x^2 + 1)dx\)

We can simplify before solving to \(\displaystyle \int_{-1}^{1}-4x^2 +4 dx\) and continue solving.

\(\displaystyle -\frac{4}{3}x^3 +4x\) from -1 to 1

\(\displaystyle -\frac{4}{3}(1)^3 + 4(1) - \left[-\frac{4}{3}(-1)^3 + 4(-1)\right]\)

\(\displaystyle -\frac{4}{3}+4 -\left[\frac{4}{3}-4\right]=-\frac{4}{3}-\frac{4}{3}+4+4 = \frac{16}{3}\)

First, figure out the boundaries of the region between the two curves. To do this, set the two functions equal to each other to determine where the curves intersect:

\(\displaystyle 5-x = \frac{6}{x}\) multiply both sides by x

\(\displaystyle 5x-x^2 = 6\)

This can be re-written in more standard quadratic form:

\(\displaystyle x^2 -5x +6 = 0\)

This can be factored as \(\displaystyle (x-2)(x-3)=0\), so our two solutions are 2 and 3. We are then evaluating the definite integral between 2 and 3. In this region, the equation \(\displaystyle 5-x\) is greater. Evaluate:

\(\displaystyle \int_{2}^{3}5-x-\frac{6}{x}dx = 5x - \frac{1}{2}x^2 -6lnx\)between 2 and 3

\(\displaystyle 5(3)-\frac{1}{2}(3)^2 - 6ln(3) - \left[5(2)-\frac{1}{2}(2)^2-6ln(2)\right]\)

\(\displaystyle 15 - 4.5 - 6.5917 - [10-2-4.1589]\approx 0.0672\)

The way to find a dot product is to multiply the first number of one vector with the first number in the other. Then you do the same with the second and third numbers in each vector. Then you add up the three values and the answer is the dot product.

\(\displaystyle (5)(-2)+(6)(3)+(-1)(9)\)

\(\displaystyle =-10+18-9\)

\(\displaystyle =-1\)