Absolute value integrals are usually evaluated by rewriting the absolute value expression as a piecewise function. If the expression inside the absolute value changes sign inside the interval, split the integral at that x-value. These problems focus on finding the vertex or zero of the absolute value expression, splitting when needed, and evaluating the resulting definite integrals.
Practice Problems
\(\textbf{1)}\) \(\displaystyle\int_{0}^{3}\left|x-2\right| \, dx\)
The answer is \(2.5\)
\(\,\,\,\,\,\,x-2=0\)
\(\,\,\,\,\,\,x=2\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3}\left|x-2\right| \, dx=\int_0^2 -(x-2)\,dx+\int_2^3(x-2)\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_0^2(2-x)\,dx+\int_2^3(x-2)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[2x-\frac{x^2}{2}\right]_0^2+\left[\frac{x^2}{2}-2x\right]_2^3\)
\(\,\,\,\,\,\,\displaystyle 2+\frac{1}{2}\)
\(\,\,\,\,\,\,\)The answer is \(2.5\)
\(\textbf{2)}\) \(\displaystyle\int_{0}^{6}\left|2x-4\right| \, dx\)
The answer is \(20\)
\(\,\,\,\,\,\,2x-4=0\)
\(\,\,\,\,\,\,x=2\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{6}\left|2x-4\right| \, dx=\int_0^2-(2x-4)\,dx+\int_2^6(2x-4)\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_0^2(4-2x)\,dx+\int_2^6(2x-4)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[4x-x^2\right]_0^2+\left[x^2-4x\right]_2^6\)
\(\,\,\,\,\,\,\displaystyle 4+16\)
\(\,\,\,\,\,\,\)The answer is \(20\)
\(\textbf{3)}\) \(\displaystyle\int_{0}^{4}\left|x+3\right| \, dx\)
The answer is \(20\)
\(\,\,\,\,\,\,x+3>0\text{ on }[0,4]\)
\(\,\,\,\,\,\,\left|x+3\right|=x+3\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{4}\left|x+3\right| \, dx=\int_0^4(x+3)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[\frac{x^2}{2}+3x\right]_0^4\)
\(\,\,\,\,\,\,\displaystyle\frac{16}{2}+12\)
\(\,\,\,\,\,\,\)The answer is \(20\)
\(\textbf{4)}\) \(\displaystyle\int_{0}^{4}\left|3x-6\right| \, dx\)
The answer is \(12\)
\(\,\,\,\,\,\,3x-6=0\)
\(\,\,\,\,\,\,x=2\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{4}\left|3x-6\right| \, dx=\int_0^2-(3x-6)\,dx+\int_2^4(3x-6)\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_0^2(6-3x)\,dx+\int_2^4(3x-6)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[6x-\frac{3x^2}{2}\right]_0^2+\left[\frac{3x^2}{2}-6x\right]_2^4\)
\(\,\,\,\,\,\,\displaystyle 6+6\)
\(\,\,\,\,\,\,\)The answer is \(12\)
\(\textbf{5)}\) \(\displaystyle\int_{-4}^{4}\left|x\right| \, dx\)
The answer is \(16\)
\(\,\,\,\,\,\,\left|x\right|\text{ is even}\)
\(\,\,\,\,\,\,\displaystyle\int_{-4}^{4}\left|x\right| \, dx=2\int_0^4x\,dx\)
\(\,\,\,\,\,\,\displaystyle 2\left[\frac{x^2}{2}\right]_0^4\)
\(\,\,\,\,\,\,\displaystyle 2\left(\frac{16}{2}\right)\)
\(\,\,\,\,\,\,\)The answer is \(16\)
\(\textbf{6)}\) \(\displaystyle\int_{0}^{3}\left|x+1\right| \, dx\)
The answer is \(7.5\)
\(\,\,\,\,\,\,x+1>0\text{ on }[0,3]\)
\(\,\,\,\,\,\,\left|x+1\right|=x+1\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3}\left|x+1\right| \, dx=\int_0^3(x+1)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[\frac{x^2}{2}+x\right]_0^3\)
\(\,\,\,\,\,\,\displaystyle\frac{9}{2}+3\)
\(\,\,\,\,\,\,\)The answer is \(7.5\)
\(\textbf{7)}\) \(\displaystyle\int_{-1}^{4}\left|3x-5\right| \, dx\)
The answer is \(\displaystyle\frac{113}{6} \approx 18.833\)
\(\,\,\,\,\,\,3x-5=0\)
\(\,\,\,\,\,\,x=\frac{5}{3}\)
\(\,\,\,\,\,\,\displaystyle\int_{-1}^{4}\left|3x-5\right| \, dx=\int_{-1}^{5/3}(5-3x)\,dx+\int_{5/3}^{4}(3x-5)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[5x-\frac{3x^2}{2}\right]_{-1}^{5/3}+\left[\frac{3x^2}{2}-5x\right]_{5/3}^{4}\)
\(\,\,\,\,\,\,\displaystyle\frac{32}{3}+\frac{49}{6}\)
\(\,\,\,\,\,\,\displaystyle\frac{64}{6}+\frac{49}{6}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{113}{6}\approx18.833\)
\(\textbf{8)}\) \(\displaystyle\int_{-1}^{2}\left|5-x\right| \, dx\)
The answer is \(13.5\)
\(\,\,\,\,\,\,5-x>0\text{ on }[-1,2]\)
\(\,\,\,\,\,\,\left|5-x\right|=5-x\)
\(\,\,\,\,\,\,\displaystyle\int_{-1}^{2}\left|5-x\right| \, dx=\int_{-1}^{2}(5-x)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[5x-\frac{x^2}{2}\right]_{-1}^{2}\)
\(\,\,\,\,\,\,\displaystyle\left(10-2\right)-\left(-5-\frac{1}{2}\right)\)
\(\,\,\,\,\,\,\)The answer is \(13.5\)
\(\textbf{9)}\) \(\displaystyle\int_{-7}^{-3}-\left|x+5\right| \, dx\)
The answer is \(-4\)
\(\,\,\,\,\,\,x+5=0\)
\(\,\,\,\,\,\,x=-5\)
\(\,\,\,\,\,\,\displaystyle\int_{-7}^{-3}-\left|x+5\right| \, dx=-\int_{-7}^{-5}-(x+5)\,dx-\int_{-5}^{-3}(x+5)\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_{-7}^{-5}(x+5)\,dx+\int_{-5}^{-3}(-x-5)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[\frac{x^2}{2}+5x\right]_{-7}^{-5}+\left[-\frac{x^2}{2}-5x\right]_{-5}^{-3}\)
\(\,\,\,\,\,\,\displaystyle -2-2\)
\(\,\,\,\,\,\,\)The answer is \(-4\)
\(\textbf{10)}\) \(\displaystyle\int_{-2}^{5}\left|x-1\right| \, dx\)
The answer is \(\displaystyle\frac{25}{2}\)
\(\,\,\,\,\,\,x-1=0\)
\(\,\,\,\,\,\,x=1\)
\(\,\,\,\,\,\,\displaystyle\int_{-2}^{5}\left|x-1\right| \, dx=\int_{-2}^{1}(1-x)\,dx+\int_1^5(x-1)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[x-\frac{x^2}{2}\right]_{-2}^{1}+\left[\frac{x^2}{2}-x\right]_1^5\)
\(\,\,\,\,\,\,\displaystyle\frac{9}{2}+8\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{25}{2}\)
\(\textbf{11)}\) \(\displaystyle\int_{-3}^{3}\left|2x\right| \, dx\)
The answer is \(18\)
\(\,\,\,\,\,\,\left|2x\right|=2\left|x\right|\)
\(\,\,\,\,\,\,\left|2x\right|\text{ is even}\)
\(\,\,\,\,\,\,\displaystyle\int_{-3}^{3}\left|2x\right| \, dx=2\int_0^3 2x\,dx\)
\(\,\,\,\,\,\,\displaystyle 4\left[\frac{x^2}{2}\right]_0^3\)
\(\,\,\,\,\,\,\displaystyle 4\left(\frac{9}{2}\right)\)
\(\,\,\,\,\,\,\)The answer is \(18\)
\(\textbf{12)}\) \(\displaystyle\int_{1}^{6}\left|x-4\right| \, dx\)
The answer is \(\displaystyle\frac{13}{2}\)
\(\,\,\,\,\,\,x-4=0\)
\(\,\,\,\,\,\,x=4\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{6}\left|x-4\right| \, dx=\int_1^4(4-x)\,dx+\int_4^6(x-4)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[4x-\frac{x^2}{2}\right]_1^4+\left[\frac{x^2}{2}-4x\right]_4^6\)
\(\,\,\,\,\,\,\displaystyle\frac{9}{2}+2\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{13}{2}\)
\(\textbf{13)}\) \(\displaystyle\int_{-2}^{2}\left|x^2-1\right| \, dx\)
The answer is \(\displaystyle\frac{8}{3}\)
\(\,\,\,\,\,\,x^2-1=0\)
\(\,\,\,\,\,\,x=-1\text{ and }x=1\)
\(\,\,\,\,\,\,\displaystyle\int_{-2}^{2}\left|x^2-1\right| \, dx=2\int_0^2\left|x^2-1\right|\,dx\)
\(\,\,\,\,\,\,\displaystyle2\left(\int_0^1(1-x^2)\,dx+\int_1^2(x^2-1)\,dx\right)\)
\(\,\,\,\,\,\,\displaystyle2\left(\left[x-\frac{x^3}{3}\right]_0^1+\left[\frac{x^3}{3}-x\right]_1^2\right)\)
\(\,\,\,\,\,\,\displaystyle2\left(\frac{2}{3}+\frac{2}{3}\right)\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{8}{3}\)
\(\textbf{14)}\) \(\displaystyle\int_{0}^{5}\left|x^2-4\right| \, dx\)
The answer is \(\displaystyle\frac{53}{3}\)
\(\,\,\,\,\,\,x^2-4=0\)
\(\,\,\,\,\,\,x=2\text{ on }[0,5]\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{5}\left|x^2-4\right| \, dx=\int_0^2(4-x^2)\,dx+\int_2^5(x^2-4)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[4x-\frac{x^3}{3}\right]_0^2+\left[\frac{x^3}{3}-4x\right]_2^5\)
\(\,\,\,\,\,\,\displaystyle\frac{16}{3}+\frac{37}{3}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{53}{3}\)
\(\textbf{15)}\) \(\displaystyle\int_{0}^{\pi}\left|\sin x\right| \, dx\)
The answer is \(2\)
\(\,\,\,\,\,\,\sin x\ge 0\text{ on }[0,\pi]\)
\(\,\,\,\,\,\,\left|\sin x\right|=\sin x\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{\pi}\left|\sin x\right| \, dx=\int_0^\pi \sin x\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[-\cos x\right]_0^\pi\)
\(\,\,\,\,\,\,\displaystyle-\cos(\pi)-\left(-\cos(0)\right)\)
\(\,\,\,\,\,\,\displaystyle1+1\)
\(\,\,\,\,\,\,\)The answer is \(2\)
\(\textbf{16)}\) \(\displaystyle\int_{0}^{2\pi}\left|\sin x\right| \, dx\)
The answer is \(4\)
\(\,\,\,\,\,\,\left|\sin x\right|\text{ has equal positive humps on }[0,2\pi]\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2\pi}\left|\sin x\right| \, dx=2\int_0^\pi \sin x\,dx\)
\(\,\,\,\,\,\,\displaystyle2\left[-\cos x\right]_0^\pi\)
\(\,\,\,\,\,\,\displaystyle2\left(2\right)\)
\(\,\,\,\,\,\,\)The answer is \(4\)
\(\textbf{17)}\) \(\displaystyle\int_{-\pi}^{\pi}\left|\cos x\right| \, dx\)
The answer is \(4\)
\(\,\,\,\,\,\,\left|\cos x\right|\text{ has four equal quarter-wave areas on }[-\pi,\pi]\)
\(\,\,\,\,\,\,\displaystyle\int_{-\pi}^{\pi}\left|\cos x\right| \, dx=4\int_0^{\pi/2}\cos x\,dx\)
\(\,\,\,\,\,\,\displaystyle4\left[\sin x\right]_0^{\pi/2}\)
\(\,\,\,\,\,\,\displaystyle4(1-0)\)
\(\,\,\,\,\,\,\)The answer is \(4\)
\(\textbf{18)}\) \(\displaystyle\int_{-3}^{3}\left|x^2-9\right| \, dx\)
The answer is \(36\)
\(\,\,\,\,\,\,x^2-9\le 0\text{ on }[-3,3]\)
\(\,\,\,\,\,\,\left|x^2-9\right|=9-x^2\)
\(\,\,\,\,\,\,\displaystyle\int_{-3}^{3}\left|x^2-9\right| \, dx=\int_{-3}^{3}(9-x^2)\,dx\)
\(\,\,\,\,\,\,\displaystyle2\int_0^3(9-x^2)\,dx\)
\(\,\,\,\,\,\,\displaystyle2\left[9x-\frac{x^3}{3}\right]_0^3\)
\(\,\,\,\,\,\,\displaystyle2(27-9)\)
\(\,\,\,\,\,\,\)The answer is \(36\)
\(\textbf{19)}\) \(\displaystyle\int_{-2}^{4}\left|2x+4\right| \, dx\)
The answer is \(36\)
\(\,\,\,\,\,\,2x+4=0\)
\(\,\,\,\,\,\,x=-2\)
\(\,\,\,\,\,\,2x+4\ge 0\text{ on }[-2,4]\)
\(\,\,\,\,\,\,\displaystyle\int_{-2}^{4}\left|2x+4\right| \, dx=\int_{-2}^{4}(2x+4)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[x^2+4x\right]_{-2}^{4}\)
\(\,\,\,\,\,\,\displaystyle(16+16)-(4-8)\)
\(\,\,\,\,\,\,\)The answer is \(36\)
\(\textbf{20)}\) \(\displaystyle\int_{-5}^{1}\left|x+2\right| \, dx\)
The answer is \(9\)
\(\,\,\,\,\,\,x+2=0\)
\(\,\,\,\,\,\,x=-2\)
\(\,\,\,\,\,\,\displaystyle\int_{-5}^{1}\left|x+2\right| \, dx=\int_{-5}^{-2}-(x+2)\,dx+\int_{-2}^{1}(x+2)\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_{-5}^{-2}(-x-2)\,dx+\int_{-2}^{1}(x+2)\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[-\frac{x^2}{2}-2x\right]_{-5}^{-2}+\left[\frac{x^2}{2}+2x\right]_{-2}^{1}\)
\(\,\,\,\,\,\,\displaystyle\frac{9}{2}+\frac{9}{2}\)
\(\,\,\,\,\,\,\)The answer is \(9\)
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