EXERCISE 6.1

 

1-4. Find the area of the shaded region.

(1)  The curve $y=f(x)=x^{2}+1$ lies above the curve $y=g(x)=x$ and the shaded region is bounded from left side at $x=-1$ and from the right side by the line $x=2$.

$$
\begin{aligned}
& A=\int_{a}^{b}[f(x)-g(x)] d x \\ \\
& A=\int_{-1}^{2}\left(x^{2}+1-x\right) d x \\ \\
& A=\left[\frac{x^{3}}{3}+x-\frac{x^{2}}{2}\right]_{-1}^{2} \\ \\
& A=\left(\frac{2^{3}}{3}+2-\frac{2^{2}}{2}\right)-\left(\frac{(-1)^{3}}{3}+(-1)-\frac{(-1)^{2}}{2}\right) \\ \\
& A=\left(\frac{8}{3}+2-\frac{4}{2}\right)-\left(-\frac{1}{3}-1-\frac{1}{2}\right) \\ \\
& A=\frac{8}{3}+2-2+\frac{1}{3}+1+\frac{1}{2}=\frac{8}{3}+\frac{1}{3}+1+\frac{1}{2} \\ \\
& A=\frac{16+2+6+3}{6}=\frac{27}{6}=\frac{9}{2}
\end{aligned}
$$

(2)   The curve $y=f(x)=\sqrt{x}$ lies above the curve $y=g(x)=-\frac{x}{4}$. The two curves intersects at $x=0$ ( the shaded region bounded from left).The shaded region is bounded by the line $x=4$ from the right side.

$$
\begin{aligned}
& A=\int_{a}^{b}[f(x)-g(x)] d x \\ \\
& A=\int_{0}^{4}\left(\sqrt{x}-\left(-\frac{x}{4}\right)\right) d x \\ \\
& A=\int_{0}^{4}\left(x^{1 / 2}+\frac{x}{4}\right) d x \\ \\
& A=\left[\frac{x^{3/2}}{ \frac{3}{2}}+\frac{x^{2}}{4(2)}\right]_{0}^{4}=\left[\frac{2}{3} x^{3 / 2}+\frac{x^{2}}{8}\right]_{0}^{4} \\ \\
& A=\left(\frac{2}{3}(4)^{3 / 2}+\frac{(4)^{2}}{8}\right)-(0) \\ \\
& A=\left(\frac{2}{3}\left(2^{2}\right)^{3 / 2}+\frac{16}{8}\right)=\frac{2}{3}(2)^{3}+\frac{16}{8}=\frac{16}{3}+\frac{16}{8} \\ \\
& A=\frac{16}{3}+2=\frac{16+6}{3}=\frac{22}{3}

\end{aligned}
$$

(3)  The shaded region is bounded by the curve $x=v(y)=\frac{1}{y^{2}}$ from the left side and by the curve $x=w(y)=y$ from the right side. Two curves intersects at $y=1$ which is the lower bound of the shaded re region and the upper bound is $y=2$.

$$
A=\int_{c}^{d}[w(y)-v(y)] d y
$$

 

$$
\begin{aligned}
& A=\int_{1}^{2}\left[y-\frac{1}{y^{2}}\right] d y=\int_{1}^{2}\left[y-y^{-2}\right] d y \\ \\
& A=\left[\frac{y^{2}}{2}-\frac{y^{-2+1}}{-2+1}\right]_{1}^{2}=\left[\frac{y^{2}}{2}-\frac{y^{-1}}{-1}\right]_{1}^{2} \\ \\
& A=\left[\frac{y^{2}}{2}+\frac{1}{y}\right]_{1}^{2}=\left(\frac{2^{2}}{2}+\frac{1}{2}\right)-\left(\frac{1}{2}+1\right) \\ \\
& A=\frac{4}{2}+\frac{1}{2}-\frac{1}{2}-1=2-1=1
\end{aligned}
$$

(4)  The shaded region is bounded by the curve $v(y)=-y$ from the left side and by the curve $\omega(y)=2-y^{2}$ from the rightside. The two curves intersects at $y=2$ which is upper bound and the lower bound is $y=0$.

$$
\begin{aligned}
& A=\int_{c}^{d}[w(y)-v(y)] d y \\ \\
& A=\int_{0}^{2}\left[2-y^{2}+y\right] d y=\left[2 y-\frac{y^{3}}{3}+\frac{y^{2}}{2}\right]_{0}^{2} \\ \\
& A=\left(2(2)-\frac{2^{3}}{3}+\frac{2^{2}}{2}\right)-(0) \\ \\
& A=4-\frac{8}{3}+\frac{4}{2}=4-\frac{8}{3}+2=6-\frac{8}{3} \\ \\
& A=\frac{18-8}{3}=\frac{10}{3}
\end{aligned}
$$

(5) Find the area of the shaded region by (a) integrating with respect to $x$ (b) integrating with respect to $y$.

(a) upper curve $$y=f(x)=2 x$$ —–(1)

Lower curve $$y=g(x)=x^{2}$$—–(2)

The point of intersection of two curves. Solving (1) and (2)

$$x^{2}=2x$$

$$x^{2}-2x=0$$ $$ \Rightarrow x(x-2)=0$$

$$x=0,$$ $$x-2=0$$

$$x=0, \quad x=2$$

Thus the two curves intersects at $x=0$ and $x=2$.

$$
\begin{aligned}
& A=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[2 \frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}=\left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2} \\ \\
& A=\left(2^{2}-\frac{2^{3}}{3}\right)-(0)=4-\frac{8}{3}=\frac{12-8}{3}=4 / 3
\end{aligned}
$$

(b) The shaded region is bounded by the curve

$
x=w(y)=\sqrt{y} \text { (because } x^{2}=y \Rightarrow x=\sqrt{y} \quad \text {as the curve lies in the first quadrant)}$ from the right side and by the curve $v(y)=y / 2$ from the left side. $y=0$ and $y=4$ are the lower and upper limits of integration These are the $y$-coordinates of point of intersection of the curves

$$
\begin{aligned}

& A=\int_{c}^{d}[w(y)-v(y)] d y=\int_{0}^{4}\left[\sqrt{y}-\frac{y}{2}\right] d y \\ \\
& A=\int_{0}^{4}\left[y^{1 / 2}-\frac{y}{2}\right] d y=\left[\frac{y^{3/2}}{3 / 2}-\frac{y^{2}}{2(2)}\right]_{0}^{4} \\ \\
& A=\left[\frac{2}{3} y^{3 / 2}-\frac{y^{2}}{4}\right]_{0}^{4}=\left(\frac{2}{3}(4)^{3 / 2}-\frac{4^{2}}{4}\right)-(0) \\ \\
& A=\frac{2}{3}(2)^{3}-4=\frac{16}{3}-4=\frac{16-12}{3}=4 / 3
\end{aligned}
$$