Answer to Question #242217 in Calculus for Josh

Question

Find the area A(R) of the region R described below.

1. R is the region bounded by y = 2x + 3, the x-axis and the line x=5 .

2. R is the region bounded by y=x^{2}+1 , the lines y=2 and y=5

3. R is the region bounded by x=y(8-y) and the y-axis.

4. R is the region bounded by y = x and y=x^{3} .

5. R is the region bounded by y=2-x^{2} and y=x^{2}-6

Answer 1

1.

$y\ =\ 2x\ +\ 3\\ x=\frac{1}{2}y-\frac{3}{2}\\ x\ intercept\ is-\frac{3}{2}\\ \int_{-\frac{3}{2}}^{5}{2x+3}\ dx\\ x^2+3x\ |_{-1.5}^5\\ 40\ -\ -2.25\ =\ 42.25\ square\ units$

2.

$y=x^2+1$

$x=\left(y-1\right)^{0.5} \\area\ of\ R=2\ \int_{2}^{5}{\left(y-1\right)^{0.5}dy}\\ 2\left(\frac{\left(y-1\right)^{1.5}}{1.5}\right)\ |_2^5\\ 2\left(\frac{4^{1.5}-1}{1.5}\right)=9.333\ square\ units$

3.

$x=y\left(8-y\right)\\ at\ the\ y\ axis,\ x=0\\ y\left(8-y\right)=0\\ y=0\ or\ y=8\\$

$\int_{0}^{8}{8y-y^2}\ dy \\4y^2-\frac{y^3}{3}\ \\4\left(64\right)-\frac{8^3}{3}=85.3333$

4.

$\\y=x \\y=x^3 \\x=x^3 \\x\left(1-x^2\right)=0 \\x=0,\ x=1,\ x=\ -1$

$Area\ of\ R=\int_{-1}^{0}{x^3-x}\ dx+\ \int_{0}^{1}{x-x^3dx}\\ \frac{x^4}{4}-\frac{x^2}{2}|_{-1}^0+\frac{x^2}{2}-\frac{x^4}{4}\ |_0^1\ \\-\frac{1}{4}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}=0.5\ square\ units$

5.

$\\y=2-x^2 \\y=x^2-6 \\2-x^2=x^2-6 \\2x^2-8=0 \\x=\ 2,\ x=\ -2$

$Area\ of\ R=\ \int_{-2}^{2}{2-x^2-\left(x^2-6\right)}dx\\ \int_{-2}^{2}{8-2x^2}\ dx\\ 8x-\frac{2}{3}x^3|_{-2}^2\\ 10\frac{2}{3}--10\frac{2}{3}=21\frac{1}{3}\ square\ units$