Chapter 8: Application of Integrals (Calculus)

In the first quadrant $y=\frac{3}{4}\sqrt{16-x^2}$, $0\le x\le 4$. By symmetry the area is 4 times the first-quadrant area:

$$A=4{\int }_0^4\frac{3}{4}\sqrt{16-x^2}dx=3{\left[\frac{x}{2}\sqrt{16-x^2}+8{\sin }^{-1}\frac{x}{4}\right]}_0^4=3\left[0+8\cdot \frac{\pi }{2}\right]=12\pi$$

12π sq units

In the first quadrant $y=\frac{3}{2}\sqrt{4-x^2}$, $0\le x\le 2$.

$$A=4{\int }_0^2\frac{3}{2}\sqrt{4-x^2}dx=6{\left[\frac{x}{2}\sqrt{4-x^2}+2{\sin }^{-1}\frac{x}{2}\right]}_0^2=6\left[2\cdot \frac{\pi }{2}\right]=6\pi$$

6π sq units

$${\int }_0^2\sqrt{4-x^2}dx={\left[\frac{x}{2}\sqrt{4-x^2}+2{\sin }^{-1}\frac{x}{2}\right]}_0^2=2\cdot \frac{\pi }{2}=\pi$$

(This is a quarter of a circle of radius 2: $\frac{1}{4}\pi (2{)}^2=\pi$.)

(A)

Use horizontal strips: $x=\frac{y^2}{4}$, from $y=0$ to $y=3$.

$$A={\int }_0^3\frac{y^2}{4}dy={\left[\frac{y^3}{12}\right]}_0^3=\frac{27}{12}=\frac{9}{4}$$

(B)

(i) ${\int }_1^2{x}^{2}dx={\left[\frac{x^3}{3}\right]}_1^2=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}$

(ii) ${\int }_1^5{x}^{4}dx={\left[\frac{x^5}{5}\right]}_1^5=\frac{3125-1}{5}=\frac{3124}{5}=624.8$

(i) 7/3 sq units (ii) 624.8 sq units

$|x+3|=-(x+3)$ for $x<-3$ and $x+3$ for $x\ge -3$. The graph is a V with vertex at (−3, 0).

$${\int }_{-6}^{-3}-(x+3)dx+{\int }_{-3}^0(x+3)dx={\left[-\frac{(x+3{)}^2}{2}\right]}_{-6}^{-3}+{\left[\frac{(x+3{)}^2}{2}\right]}_{-3}^0=\frac{9}{2}+\frac{9}{2}$$

(Two triangles, each of base 3 and height 3.)

9

$\sin x\ge 0$ on $[0,\pi ]$ and $\le 0$ on $[\pi ,2\pi ]$, so

$$A={\int }_0^{\pi }\sin xdx+\left|{\int }_{\pi }^{2\pi }\sin xdx\right|=[-\cos x{]}_0^{\pi }+|[-\cos x{]}_{\pi }^{2\pi }|=2+|-2|=4$$

4 sq units

$x^3<0$ on $(-2,0)$, so

$$A=\left|{\int }_{-2}^0{x}^{3}dx\right|+{\int }_0^1{x}^{3}dx=\left|{\left[\frac{x^4}{4}\right]}_{-2}^0\right|+\frac{1}{4}=4+\frac{1}{4}=\frac{17}{4}$$

(D)

The curve is symmetric about the origin, and each half has area ${\int }_0^1{x}^{2}dx=\frac{1}{3}$.

$A=\left|{\int }_{-1}^0(-x^2)dx\right|+{\int }_0^1{x}^{2}dx=\frac{1}{3}+\frac{1}{3}$

(C)