EXERCISE 6.1

EXERCISE 6.1

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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

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