Exercise 21
Find the area bounded by the lines and .
Matches the book, not yet reviewed
This working reaches the answer NCERT prints. It has not yet been read through by hand.
NCERT’s answer
\(\displaystyle \frac{15}{2}\) sq. units
Pairwise intersections of the lines give the vertices.\[y=4x+5,\ y=5-x \Rightarrow 5x=0 \Rightarrow A(0,5) \]
\[y=4x+5,\ 4y=x+5 \Rightarrow 16x+20=x+5 \Rightarrow B(-1,1) \]
\[y=5-x,\ 4y=x+5 \Rightarrow 20-4x=x+5 \Rightarrow C(3,2) \]
Above \(\displaystyle 4y=x+5\), the upper boundary is \(\displaystyle y=4x+5\) for \(\displaystyle -1\le x\le0\) and \(\displaystyle y=5-x\) for \(\displaystyle 0\le x\le3\).\[\text{Area}=\int_{-1}^{0}\left[(4x+5)-\frac{x+5}{4}\right]dx+\int_0^3\left[(5-x)-\frac{x+5}{4}\right]dx \]
\[=\int_{-1}^{0}\frac{15}{4}(x+1)\,dx+\int_0^3\frac54(3-x)\,dx \]
\[=\frac{15}{8}\Big[(x+1)^2\Big]_{-1}^{0}-\frac58\Big[(3-x)^2\Big]_0^3 \]
\[=\frac{15}{8}+\frac{45}{8}=\frac{15}{2} \]Answer: \(\displaystyle \dfrac{15}{2}\) square units
Above \(\displaystyle 4y=x+5\), the upper boundary is \(\displaystyle y=4x+5\) for \(\displaystyle -1\le x\le0\) and \(\displaystyle y=5-x\) for \(\displaystyle 0\le x\le3\).\[\text{Area}=\int_{-1}^{0}\left[(4x+5)-\frac{x+5}{4}\right]dx+\int_0^3\left[(5-x)-\frac{x+5}{4}\right]dx \]
\[=\int_{-1}^{0}\frac{15}{4}(x+1)\,dx+\int_0^3\frac54(3-x)\,dx \]
\[=\frac{15}{8}\Big[(x+1)^2\Big]_{-1}^{0}-\frac58\Big[(3-x)^2\Big]_0^3 \]
\[=\frac{15}{8}+\frac{45}{8}=\frac{15}{2} \]Answer: \(\displaystyle \dfrac{15}{2}\) square units
\[A=\int_0^{\pi/2}2\cos x\,dx+\left|\int_{\pi/2}^{3\pi/2}2\cos x\,dx\right|+\int_{3\pi/2}^{2\pi}2\cos x\,dx \]\[\int_0^{\pi/2}2\cos x\,dx=2\Big[\sin x\Big]_0^{\pi/2}=2 \]\[\int_{\pi/2}^{3\pi/2}2\cos x\,dx=2\Big[\sin x\Big]_{\pi/2}^{3\pi/2}=2(-1-1)=-4 \]\[\int_{3\pi/2}^{2\pi}2\cos x\,dx=2\Big[\sin x\Big]_{3\pi/2}^{2\pi}=2(0+1)=2 \]\[A=2+|-4|+2=8 \]Answer: \(\displaystyle 8\) square units.
\[A=\int_{-3}^{-1}(-x)\,dx+\int_{-1}^{3}(x+2)\,dx \]\[\int_{-3}^{-1}(-x)\,dx=\left[-\frac{x^2}{2}\right]_{-3}^{-1}=-\frac12+\frac92=4 \]\[\int_{-1}^{3}(x+2)\,dx=\left[\frac{x^2}{2}+2x\right]_{-1}^{3}=\frac{21}{2}-\left(-\frac32\right)=12 \]\[A=4+12=16 \]Answer: \(\displaystyle 16\) square units.
\[\sin x=\cos x \;\Rightarrow\; x=\frac{\pi}{4} \]\[A=\int_0^{\pi/4}(\cos x-\sin x)\,dx=\Big[\sin x+\cos x\Big]_0^{\pi/4} \]\[A=\left(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\right)-(0+1)=\sqrt2-1 \]
On \(\displaystyle [-1,2]\) the line lies above the parabola.\[A=\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^2}{4}\right)dx=\frac14\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2} \]\[A=\frac14\left[\left(2+4-\frac83\right)-\left(\frac12-2+\frac13\right)\right]=\frac14\left(\frac{10}{3}+\frac76\right)=\frac14\cdot\frac92=\frac98 \]
\[A=2\int_0^4\sqrt{16-x^2}\,dx=2\left[\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\frac{x}{4}\right]_0^4 \]\[A=2\left(0+8\cdot\frac{\pi}{2}\right)=8\pi \]
Under the line up to \(\displaystyle x=4\), then under the circle up to \(\displaystyle x=4\sqrt2\):\[A=\int_0^4 x\,dx+\int_4^{4\sqrt2}\sqrt{32-x^2}\,dx \]\[\int_0^4 x\,dx=8 \]\[\int_4^{4\sqrt2}\sqrt{32-x^2}\,dx=\left[\frac{x}{2}\sqrt{32-x^2}+16\sin^{-1}\frac{x}{4\sqrt2}\right]_4^{4\sqrt2}=8\pi-(8+4\pi)=4\pi-8 \]\[A=8+4\pi-8=4\pi \]
\[A=\int_0^{\pi/2}\cos x\,dx+\left|\int_{\pi/2}^{\pi}\cos x\,dx\right| \]\[A=\Big[\sin x\Big]_0^{\pi/2}+\left|\Big[\sin x\Big]_{\pi/2}^{\pi}\right|=1+|0-1|=2 \]
On \(\displaystyle [0,4]\) the parabola lies above the line:\[A = \int_0^4 \left(\sqrt{x} - \frac{x}{2}\right)dx \]\[= \left[\frac{2}{3}x^{3/2} - \frac{x^2}{4}\right]_0^4 \]\[= \frac{16}{3} - 4 = \frac{4}{3} \]
\[A=\int_0^{\pi/2}\sin x\,dx \]
\[=\left[-\cos x\right]_0^{\pi/2} \]
\[=-\cos\frac{\pi}{2}+\cos 0=0+1=1 \]