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Chapter 5: Problem 72

Use geometry to evaluate the following integrals. $$\int_{-2}^{3}|x+1| d x$$

Short Answer

Expert verified
Answer: The function inside the absolute value changes its sign at \(x = -1\), and the value of the integral \(\int_{-2}^{3}|x+1| dx\) is 8.

Step by step solution

01

Identify where the function inside the absolute value changes its sign

We have the function inside the absolute value as \(x+1\). We need to find the point where \(x+1 = 0\). Solve for x: $$x+1 = 0$$ $$x = -1$$ Thus, the function inside the absolute value changes its sign at \(x = -1\).
02

Break the integral into parts

Now that we know the point where the function changes its sign, we can split the integral into separate parts (considering the interval \([-2, 3]\)): $$\int_{-2}^{3}|x+1| dx = \int_{-2}^{-1}|x+1| dx + \int_{-1}^{3}|x+1| dx$$ For the first part of the integral, between \(-2\) and \(-1\), we have \(x+1 < 0\). Therefore, we can rewrite the integral as: $$\int_{-2}^{-1}|x+1| dx = -\int_{-2}^{-1}(x+1) dx$$ For the second part of the integral, between \(-1\) and \(3\), we have \(x+1 > 0\). Therefore, we can rewrite the integral as: $$\int_{-1}^{3}|x+1| dx = \int_{-1}^{3}(x+1) dx$$ Now, the original integral becomes: $$\int_{-2}^{3}|x+1| dx = -\int_{-2}^{-1}(x+1) dx + \int_{-1}^{3}(x+1) dx$$
03

Evaluate each integral separately

Now, we will evaluate both integrals one by one. The first integral: $$-\int_{-2}^{-1}(x+1) dx$$ $$=-\left[\frac{1}{2}x^2+x\right]_{-2}^{-1}$$ $$=-\left(\frac{1}{2}(-1)^2+(-1)-\frac{1}{2}(-2)^2+(-2)\right)$$ $$=-\left(\frac{1}{2}-1-\frac{1}{2}(-2)\right)$$ $$=-\left(-\frac{5}{2}\right)$$ $$=\frac{5}{2}$$ The second integral: $$\int_{-1}^{3}(x+1) dx$$ $$=\left[\frac{1}{2}x^2+x\right]_{-1}^{3}$$ $$=\left(\frac{1}{2}(3)^2+(3)-\frac{1}{2}(-1)^2+(-1)\right)$$ $$=\left(\frac{9}{2}+3-\frac{1}{2}-1\right)$$ $$=\frac{11}{2}$$
04

Add the results of both integrals

Finally, put the values of both integrals back together and add them up: $$\int_{-2}^{3}|x+1| dx = \frac{5}{2}+\frac{11}{2}$$ $$\int_{-2 }^{3}|x+1| dx = \frac{16}{2}$$ $$\int_{-2}^{3}|x+1| dx = 8$$ So, the value of the integral is 8.

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