Chapter 8: Application of Integrals

✅ Solution By symmetry, Area \(= 4\displaystyle\int_0^a \sqrt{a^2-x^2}\,dx\). Using formula \(\displaystyle\int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C\): \[= 4\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a = 4\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi a^2\]

✅ Solution Area \(= 4\displaystyle\int_0^a \frac{b}{a}\sqrt{a^2-x^2}\,dx = \frac{4b}{a}\cdot\frac{\pi a^2}{4} = \pi ab\).

✅ Solution This is the area of one quarter of the circle \(x^2+y^2=4\) (radius \(=2\)). Area \(=\dfrac{1}{4}\pi(2)^2 = \pi\).

✅ Solution From \(y^2=4x\), we get \(x=\dfrac{y^2}{4}\). Area \(=\displaystyle\int_0^3 \frac{y^2}{4}\,dy = \left[\frac{y^3}{12}\right]_0^3 = \frac{27}{12} = \frac{9}{4}\).

✅ Solution When taking vertical strips, each strip has height \(y\) and width \(dx\). So the elementary area is \(dA = y\,dx\). The total area is \(A = \displaystyle\int_a^b y\,dx = \int_a^b f(x)\,dx\).

✅ Solution \(y = x^3 \leq 0\) on \([-2, 0]\) and \(\geq 0\) on \([0, 1]\). \[A = \left|\int_{-2}^0 x^3\,dx\right| + \int_0^1 x^3\,dx = \left|\left[\frac{x^4}{4}\right]_{-2}^0\right| + \left[\frac{x^4}{4}\right]_0^1 = \left|-\frac{16}{4}\right| + \frac{1}{4} = 4 + \frac{1}{4} = \frac{17}{4}\]

✅ Solution Here \(a=4,\, b=3\). Area of ellipse \(= \pi ab = \pi \times 4 \times 3 = 12\pi\).

✅ Solution Here \(a=2,\, b=3\). Area \(= \pi ab = \pi \times 2 \times 3 = 6\pi\).

✅ Solution When \(f(x) < 0\), the integral \(\displaystyle\int_a^b f(x)\,dx\) is negative. But area is always positive, so we take the absolute value: Area \(= \left|\displaystyle\int_a^b f(x)\,dx\right|\).

✅ Solution Split over regions where \(\cos x\) is positive/negative: \[A = \int_0^{\pi/2}\cos x\,dx + \left|\int_{\pi/2}^{3\pi/2}\cos x\,dx\right| + \int_{3\pi/2}^{2\pi}\cos x\,dx = 1 + 2 + 1 = 4\]

✅ Solution Note: \(y = -x^2\) for \(x<0\) and \(y = x^2\) for \(x>0\). \[A = \left|\int_{-1}^0 -x^2\,dx\right| + \int_0^1 x^2\,dx = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}\]

✅ Solution \(\displaystyle\int_1^2 x^2\,dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\).

✅ Solution When the curve is expressed as \(x = g(y)\), we use horizontal strips of width \(dy\) and length \(x = g(y)\). Area \(= \displaystyle\int_c^d x\,dy = \int_c^d g(y)\,dy\).

✅ Solution \(\displaystyle\int_1^5 x^4\,dx = \left[\frac{x^5}{5}\right]_1^5 = \frac{3125}{5} - \frac{1}{5} = \frac{3124}{5}\).

✅ Solution Setting \(y = 0\): \(3x + 2 = 0 \Rightarrow x = -\dfrac{2}{3}\).

✅ Solution \(\displaystyle\int_1^2 x^2\,dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\).

✅ Solution \(\sin x \geq 0\) on \([0,\pi]\) and \(\leq 0\) on \([\pi,2\pi]\). \[A = \int_0^\pi \sin x\,dx + \left|\int_\pi^{2\pi}\sin x\,dx\right| = [-\cos x]_0^\pi + |[-\cos x]_\pi^{2\pi}| = 2 + 2 = 4\]

✅ Solution Radius \(a = 3\). Area \(= \pi a^2 = \pi \times 9 = 9\pi\).

✅ Solution The line crosses x-axis at \(x = -\dfrac{2}{3}\). Split integral: \[A = \left|\int_{-1}^{-2/3}(3x+2)\,dx\right| + \int_{-2/3}^{1}(3x+2)\,dx = \frac{1}{6} + \frac{25}{6} = \frac{26}{6} = \frac{13}{3}\]

✅ Solution Total area of ellipse \(= \pi ab = \pi \times 5 \times 4 = 20\pi\). First quadrant area \(= \dfrac{1}{4} \times 20\pi = 5\pi\).

✅ Solution From \(y^2 = 4x \Rightarrow x = \dfrac{y^2}{4}\). Area \(= \displaystyle\int_0^3 \frac{y^2}{4}\,dy = \frac{1}{4}\left[\frac{y^3}{3}\right]_0^3 = \frac{1}{4} \cdot \frac{27}{3} = \frac{9}{4}\).

✅ Solution \(x+3 = 0 \Rightarrow x=-3\). Split: \[\int_{-6}^{-3}-(x+3)\,dx + \int_{-3}^{0}(x+3)\,dx = \left[-\frac{(x+3)^2}{2}\right]_{-6}^{-3} + \left[\frac{(x+3)^2}{2}\right]_{-3}^{0} = \frac{9}{2} + \frac{9}{2} = 9\]

✅ Solution Area of full circle \(= \pi(2)^2 = 4\pi\). Semicircle area \(= \dfrac{4\pi}{2} = 2\pi\).
Alternatively: \(\displaystyle\int_{-2}^2\sqrt{4-x^2}\,dx = \frac{\pi(2)^2}{2} = 2\pi\).

✅ Solution This is the first-quadrant area of circle of radius 2: \[\displaystyle\int_0^2\sqrt{4-x^2}\,dx = \left[\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\frac{x}{2}\right]_0^2 = 0 + 2\cdot\frac{\pi}{2} = \pi\]

✅ Solution Here \(a^2=9 \Rightarrow a=3\) and \(b^2=4 \Rightarrow b=2\). Area \(= \pi ab = \pi \times 3 \times 2 = 6\pi\).