Question

evaluate the integral by interpreting it in terms of areas. in other words, draw a picture of the region the integral represents, and find the area using high school geometry.
\\(\int_{0}^{8}|8x - 3|dx\\)

Explanation:

Step1: Find where the expression inside absolute - value is zero

Set $8x - 3=0$, then $x=\frac{3}{8}$.

Step2: Split the integral based on the zero - point

We split $\int_{0}^{8}|8x - 3|dx$ into two integrals: $\int_{0}^{\frac{3}{8}}-(8x - 3)dx+\int_{\frac{3}{8}}^{8}(8x - 3)dx$.

Step3: Evaluate the first integral

For $\int_{0}^{\frac{3}{8}}-(8x - 3)dx=\int_{0}^{\frac{3}{8}}(- 8x + 3)dx$. Using the power rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)$, we have $[-4x^{2}+3x]_0^{\frac{3}{8}}=-4(\frac{3}{8})^{2}+3(\frac{3}{8})-(-4(0)^{2}+3(0))=-4\times\frac{9}{64}+\frac{9}{8}=-\frac{9}{16}+\frac{9}{8}=\frac{-9 + 18}{16}=\frac{9}{16}$.

Step4: Evaluate the second integral

For $\int_{\frac{3}{8}}^{8}(8x - 3)dx=[4x^{2}-3x]_{\frac{3}{8}}^{8}=4(8)^{2}-3(8)-(4(\frac{3}{8})^{2}-3(\frac{3}{8}))=4\times64-24-(4\times\frac{9}{64}-\frac{9}{8})=256-24-(\frac{9}{16}-\frac{9}{8})=232-(\frac{9 - 18}{16})=232+\frac{9}{16}=\frac{3712+9}{16}=\frac{3721}{16}$.

Step5: Sum the results of the two integrals

$\frac{9}{16}+\frac{3721}{16}=\frac{9 + 3721}{16}=\frac{3730}{16}=\frac{1865}{8}=233.125$.

In terms of geometry, the function $y = |8x - 3|$ is a V - shaped function. The integral $\int_{0}^{8}|8x - 3|dx$ represents the area between the curve $y = |8x - 3|$, the $x$ - axis, $x = 0$, and $x = 8$. The area is composed of two right - angled triangles (after splitting based on the vertex of the V - shaped function at $x=\frac{3}{8}$).

Answer:

$\frac{1865}{8}$