CALCULUS: Find Integral from -8 to -2 by Interpreting in Terms of Areas

[MarkFL]

Here is the question:

CALCULUS: Find the integral by interpreting the integral in terms of sums and/or differences of areas...?


You are given the four points in the plane A=(−8,1), B=(−6,−8),
C=(−4,2), and D=(−2,−3). The graph of the function f(x) consists of the three line segments AB, BC and CD. Find the integral f(x)dx from−8 to−2 by interpreting the integral in terms of sums and/or differences of areas of elementary figures.

The integral f(x)dx from -8 to -2 = ?

I have posted a link there to this thread so the OP can see my work.

 

[MarkFL]

Hello Rankel,

Let's plot the given points and connect them with line segments:

View attachment 1728

As you can see, we have four triangles, two above the $x$-axis, and two below. Beginning on the left, we see the first triangle above the $x$-axis has a height $h$ of 1. Its base $b$ can be found by similarity:

\(\displaystyle \frac{b}{1}=\frac{2}{9}\)

Hence, its area is:

\(\displaystyle A_1=\frac{1}{2}bh=\frac{1}{2}\cdot\frac{2}{9}\cdot1=\frac{1}{9}\)

Moving to the right to the first triangle under the $x$-axis, we see its height is 8. From the first triangle, we know the $x$-coordinate of the left end of its base is:

\(\displaystyle x_1=-8+\frac{2}{9}=-\frac{70}{9}\)

Using similarity, we find the $x$-coordinate of the right end of the base is:

\(\displaystyle x_2=-4-\frac{2}{5}=-\frac{22}{5}\)

Hence the base is:

\(\displaystyle b=-\frac{22}{5}+\frac{70}{9}=\frac{152}{45}\)

And so the area of the second triangle is given by:

\(\displaystyle A_2=\frac{1}{2}bh=\frac{1}{2}\cdot\frac{152}{45} \cdot8=\frac{608}{45}\)

Next, moving to the right to the second triangle above the $x$-axis, we see its height is 2. We know the left end of its base is:

\(\displaystyle x_1=-\frac{22}{5}\)

And, using similarity, we find the right end of the base is at:

\(\displaystyle x_2=-4+\frac{4}{5}=-\frac{16}{5}\)

Hence the length of the base is:

\(\displaystyle b=-\frac{16}{5}+\frac{22}{5}=\frac{6}{5}\)

And so the area of the third triangle is:

\(\displaystyle A_3=\frac{1}{2}bh=\frac{1}{2}\cdot\frac{6}{5}\cdot2=\frac{6}{5}\)

And moving to the second triangle below the $x$-axis, we see its height is 3 and its base is:

\(\displaystyle b=-2+\frac{16}{5}=\frac{6}{5}\)

Thus, its area is:

\(\displaystyle A_4=\frac{1}{2}bh=\frac{1}{2}\cdot\frac{6}{5}\cdot3=\frac{9}{5}\)

So, we may now state:

\(\displaystyle \int_{-8}^{-2} f(x)\,dx=A_1-A_2+A_3-A_4=\frac{1}{9}-\frac{608}{45}+\frac{6}{5}-\frac{9}{5}=-14\)

As a check, we may find the linear equations describing the lines segments and then integrate over the appropriate intervals. The line passing through the points $\left(x_1,y_1 \right)$ and $\left(x_2,y_2 \right)$ is given by:

\(\displaystyle y=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1 \right)+y_1\)

Thus, we find the following:

\(\displaystyle \overline{AB}\implies y=\frac{-8-1}{-6+8}(x+8)+1=-\frac{9}{2}x-35\)

\(\displaystyle \overline{BC}\implies y=\frac{2+8}{-4+6}(x+6)-8=5x+22\)

\(\displaystyle \overline{CD}\implies y=\frac{-3-2}{-2+4}(x+4)+2=-\frac{5}{2}x-8\)

And so we have:

\(\displaystyle \int_{-8}^{-2} f(x)\,dx=\int_{-8}^{-6} -\frac{9}{2}x-35\,dx+\int_{-6}^{-4} 5x+22\,dx+\int_{-4}^{-2} -\frac{5}{2}x-8\,dx\)

\(\displaystyle \int_{-8}^{-6} -\frac{9}{2}x-35\,dx=-\left[\frac{9}{4}x^2+35x \right]_{-8}^{-6}=\)

\(\displaystyle -\left(\left(\frac{9}{4}(-6)^2+35(-6) \right)-\left(\frac{9}{4}(-8)^2+35(-8) \right) \right)=-\left(-129+136 \right)=-7\)

\(\displaystyle \int_{-6}^{-4} 5x+22\,dx=\left[\frac{5}{2}x^2+22x \right]{-6}^{-4}=\)

\(\displaystyle \left(\frac{5}{2}(-4)^2+22(-4) \right)-\left(\frac{5}{2}(-6)^2+22(-6) \right)=-48+42=-6\)

\(\displaystyle \int_{-4}^{-2} -\frac{5}{2}x-8\,dx=-\left[\frac{5}{4}x^2+8x \right]{-4}^{-2}=\)

\(\displaystyle -\left(\left(\frac{5}{4}(-2)^2+8(-2) \right)-\left(\frac{5}{4}(-4)^2+8(-4) \right) \right)=-\left(-11+12 \right)=-1\)

And so we find:

\(\displaystyle \int_{-8}^{-2} f(x)\,dx=-7-6-1=-14\)

And this checks with our previous result.