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Chapter 5: Problem 44

Use symmetry to evaluate the following integrals. $$\int_{-2}^{2}\left(1-|x|^{3}\right) d x$$

Short Answer

Expert verified
Based on the step-by-step solution above, create a short answer to the question: To evaluate the definite integral, first recognize that the given function is an even function due to the even exponent inside the absolute value function. Since it is an even function, we can apply the symmetry property and rewrite the integral as $$2\int_{0}^{2} (1-|x|^3) dx$$. Simplify the integral to $$2\int_{0}^{2} (1-x^3) dx$$ and evaluate it. The final answer for the given definite integral is $$-4$$.

Step by step solution

01

Observe the given function

Consider the given function, $$f(x) = 1 - |x|^3$$ Notice that \(f(x)\) has an even exponent inside the absolute value function, which means that \(f(x)\) is an even function. An even function satisfies the property: $$f(-x) = f(x)$$
02

Utilize symmetry property

Since \(f(x)\) is an even function, we can apply symmetry property in the given integral: $$\int_{-2}^{2} (1-|x|^3) dx = 2\int_{0}^{2} (1-|x|^3) dx$$ We have utilized the fact that the area under the graph of an even function between \(-a\) and \(a\) is twice the area under the graph between \(0\) and \(a\).
03

Calculate the definite integral

Now, let's evaluate the definite integral we got after applying symmetry property: $$2\int_{0}^{2} (1-|x|^3) dx$$ Since \(x\) is positive in the interval \([0, 2]\), the absolute value of \(x^3\) will be \(x^3\). $$2\int_{0}^{2} (1-x^3) dx$$ Now we will integrate the function with respect to \(x\) and apply the limits of the integral: $$2\left[\int(1)dx - \int(x^3)dx \right]_0^2$$ $$2\left[(x - \frac{x^4}{4})\right]_0^2$$ $$2\left[(2 - \frac{2^4}{4}) - (0 - \frac{0^4}{4})\right]$$ $$2\left[2 - \frac{16}{4}\right]$$ $$2\left[2 - 4\right]$$ $$2 (-2)$$
04

Final answer

The final answer for the given definite integral is: $$\int_{-2}^{2}\left(1-|x|^{3}\right) d x = -4$$

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Most popular questions from this chapter

a. Consider the curve \(y=1 / x,\) for \(x \geq 1\). For what value of \(b>0\) does the region bounded by this curve and the \(x\) -axis on the interval \([1, b]\) have an area of \(1 ?\) b. Consider the curve \(y=1 / x^{p},\) where \(x \geq 1,\) and \(p<2\) with \(p \neq 1 .\) For what value of \(b\) (as a function of \(p\) ) does the region bounded by this curve and the \(x\) -axis on the interval \([1, b]\) have unit area? c. Is \(b(p)\) in part (b) an increasing or decreasing function of \(p ?\) Explain.

Consider the integral \(I=\int_{0}^{\pi / 2} \sin x d x\), a. Write the left Riemann sum for \(I\) with \(n\) subintervals. b. Show that \(\lim _{\theta \rightarrow 0} \theta\left(\frac{\cos \theta+\sin \theta-1}{2(1-\cos \theta)}\right)=1\). c. It is a fact that \(\sum_{k=0}^{n-1} \sin \left(\frac{\pi k}{2 n}\right)=\frac{\cos \left(\frac{\pi}{2 n}\right)+\sin \left(\frac{\pi}{2 n}\right)-1}{2\left(1-\cos \left(\frac{\pi}{2 n}\right)\right)}\). Use this fact and part (b) to evaluate \(I\) by taking the limit of the Riemann sum as \(n \rightarrow \infty\).

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