Answer to Question #172135 in Calculus for Ceejay

Question

A. Find the area underneath the given curve.

1.) y= -2x²+8

from x=0 to x=1

2.) y= -x²+7

from x=0 to x=1

B. Find the area enclosed by the given curve, the x-axis , and the given lines.

1.) y=-1/3 x² + 10,

from x=1 and x=3

2.) y= x³-8

from x=-1 to x=2

C. Find the area bounded by the given curve and line.

1.) y= x² + 3 and y=7

2.) y= 2x² and y= 4x + 6

3.) y = -x² + 4x and y= x²

4.) y= x³ - 6x² + 8x and y= x² - 4x

Answer 1

A. 1) $\int_{0}^{1}{-2x^2+8\ dx}$

$-\frac{2}{3}x^3+8x$

$-\frac{2}{3}+8-0=7\frac{1}{3}$

2) $\int_{0}^{1}{-x^2+1\ dx}$

$-\frac{x^3}{3}+x$

$-\frac{1}{3}+1-0=\frac{2}{3}$

B. 1) $\int_{1}^{3}{-\frac{1}{3}x^2+10\ dx}$

$-\frac{x^3}{9}+10x$

$-\frac{27}{9}+30-\left(-\frac{1}{9}+10\right)=17\frac{1}{9}$

2) $\int_{-1}^{2}{x^3-8\ dx}$

$\frac{x^4}{4}-8x$

$\frac{16}{4}-16-\left(\frac{1}{4}-\ -8\right)=\ -20\frac{1}{4}=20\frac{1}{4}\ square\ units$

C. 1) $y=x^2+3$

$y=7$

$7=x^2+3=>x=\ \pm2$

$\int_{-2}^{2}{7-\left(x^2+3\ \right)dx}$

$4x-\frac{x^3}{3}$

$8-\frac{8}{3}-\left(-8-\ -\frac{8}{3}\right)=10\frac{2}{3}$

2) $y=2x^2$

$y=4x+6$

$2x^2=4x+6$

$x^2-2x-3=0$

$x^2-3x+x-3=0$

$x\left(x-3\right)+\left(x-3\right)=0$

$\left(x+1\right)\left(x-3\right)=0$

$x=\ -1\ or\ x=3$

$\int_{-1}^{3}{4x+6-2x^2\ dx}$

$2x^2+6x-\frac{2}{3}x^3$

$18+18-\frac{54}{3}-\left(2-6--\frac{2}{3}\right)=21\frac{1}{3}$

3) y=\ -x^2+4x\

$y=x^2$

x^2=\ -x^2+4x\

$2x^2-4x=0$

$x\left(x-2\right)=0$

$x=0\ or\ x=2$

$\int_{0}^{2}{-x^2+4x-x^2\ dx}$

$\int_{0}^{2}{-x^2+4x-x^2\ dx}$

$-\frac{2}{3}x^3+2x^2$

$-\frac{16}{3}+8=2\frac{2}{3}$

4) $y=x^3-6x^2$

$y=x^2-4x$

$x^3-7x^2+4x=0$

$x\left(x^2-7x+4\right)=0=>x=0$

$x^2-7x+4=0$

$\frac{7\ \pm\ \sqrt{49-16}}{2}=>x=\ 0.628\ or\ 6.372$

$x\ intercepts$

${for\ y=x^2-4x\ ,\ \ x}^2-4x=0=>x=0\ or\ x=4$

$for\ y=\ x^3-6x^2, \ x^3-6x^2=0=>x=0\ or\ x=6$

$|\int_{0}^{6}{x^3-6x^2}dx|-|\int_{0}^{4}{x^2-4x\ dx\ |}$

$+\ \int_{4}^{6.372}{x^2-4x\ dx\ }-\ \int_{6}^{6.372}{x^3-6x^2\ dx\ }$

$\left|\frac{x^4}{4}-2x^3\right|0,6-\left|\frac{x^3}{3}-2x^2\right|0,\ 4$

$+\left(\frac{x^3}{3}-2x^2\right),\ 4,\ 6.372-\left(\frac{x^4}{4}-2x^3\right),\ 6,\ 6.372$

$108-10.667+15.701-\ 2.702=110.332$