Answer:
Areas under the curve
Step-by-step explanation:
The area under a curve on an interval [a, b] is the integral of the function computed in this interval : [tex]A=\int _a^b|f\left(x\right)|dx[/tex]
(1) For [tex]f\left(x\right)=3\sqrt{x}[/tex] with [tex]a=1,\:b=4[/tex]
area [tex]=\int _1^4\left3\sqrt{x}\;dx[/tex] = [tex]3\cdot \int _1^4\sqrt{x}dx[/tex] =
[tex]\int \sqrt{x}[/tex] = [tex]\frac{2}{3}x^{\frac{3}{2}}[/tex]
[tex]3\cdot\frac{2}{3}x^{\frac{3}{2}}[/tex] = [tex]2x^{\frac{3}{2}}[/tex]
At [tex]x = 4,[/tex] we get [tex]2\cdot4^{\frac{3}{2}}[/tex] = [tex]2\cdot8 = 16[/tex]
At [tex]x = 1,[/tex] we get [tex]2\cdot1^{\frac{3}{2}}[/tex] = [tex]2.1 = 2[/tex]
So area under the curve for [tex]f(x) = \:3\sqrt{x}[/tex] in the interval [tex][1, 4] = 14[/tex]
(2) [tex]f\left(x\right)=x-\frac{2}{3}\\[/tex]
[tex]\int \:x-\frac{2}{3}dx[/tex] = [tex]\int \:xdx-\int \frac{2}{3}dx[/tex] [tex]=\frac{x^2}{2}-\frac{2}{3}x[/tex]
[tex]\left[\frac{x^2}{2}\right]^{27}_1 = \frac{27^2}{2} - \frac{1}{2} = \frac{729}{2}-\frac{1}{2} = \frac{728}{2} = 364[/tex]
[tex]\left[\frac{2}{3}x\right]^{27}_1 = \frac{2}{3}\cdot \:27 - \frac{2}{3}\cdot \:1 = 18-\frac{2}{3} = \frac{52}{3}[/tex]
[tex]\int _1^{27}\left|x-\frac{2}{3}\right|dx = 364-\frac{52}{3} = \frac{1040}{3}[/tex] (Answer)
