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Chapter 7: Problem 46

Evaluate the following integrals or state that they diverge. $$\int_{1}^{11} \frac{d x}{(x-3)^{2 / 3}}$$

Short Answer

Expert verified
Question: Evaluate the definite integral $$\int_{1}^{11} \frac{d x}{(x-3)^{2 / 3}}$$ Answer: $$6 - 3 \cdot (-2)^{1/3}$$

Step by step solution

01

Define the substitution

Let's make the substitution: $$u = x - 3$$ Then, the derivative of u with respect to x is: $$\frac{d u}{d x} = 1$$ Now, we can rewrite the integral in terms of u, and we also need to change the limits of integration.
02

Rewrite the integral in terms of u

First, let's find the new limits. When x = 1, $$u = 1 - 3 = -2$$ When x = 11, $$u = 11 - 3 = 8$$ Now, we change the integral to have u as the variable. Since \(\frac{d u}{d x} = 1\), we have: $$\int_{-2}^{8} \frac{d u}{u^{2 / 3}}$$
03

Find the anti-derivative

Now, we find the anti-derivative of the function \(\frac{1}{u^{2 / 3}}\) with respect to u. We have: $$\int \frac{d u}{u^{2 / 3}} = \int u^{-2/3} d u$$ Applying the power rule, we get: $$\frac{u^{1/3}}{\frac{1}{3}} + C = 3u^{1/3} + C$$
04

Apply the limits and evaluate the definite integral

Now that we have the anti-derivative, we can evaluate the definite integral. We have: $$\left(3u^{1/3} + C\right) \Big|_{-2}^{8} = 3 \cdot 8^{1/3} - 3 \cdot (-2)^{1/3}$$ Now evaluate, $$3 \cdot 2 - 3 \cdot (-2)^{1/3} = 6 - 3 \cdot (-2)^{1/3}$$ Hence, the value of the integral is $$\int_{1}^{11} \frac{d x}{(x-3)^{2 / 3}} = 6 - 3 \cdot (-2)^{1/3}$$

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Most popular questions from this chapter

Powers of sine and cosine It can be shown that \(\int_{0}^{\pi / 2} \sin ^{n} x d x=\int_{0}^{\pi / 2} \cos ^{n} x d x=\) \(\left\\{\begin{array}{ll}\frac{1 \cdot 3 \cdot 5 \cdots(n-1)}{2 \cdot 4 \cdot 6 \cdots n} \cdot \frac{\pi}{2} & \text { if } n \geq 2 \text { is an even integer } \\ \frac{2 \cdot 4 \cdot 6 \cdots(n-1)}{3 \cdot 5 \cdot 7 \cdots n} & \text { if } n \geq 3 \text { is an odd integer. }\end{array}\right.\)

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Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=(x-1)^{-1 / 4}\) and the \(x\) -axis on the interval (1,2] is revolved about the \(x\) -axis.

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A total charge of \(Q\) is distributed uniformly on a line segment of length \(2 L\) along the \(y\) -axis (see figure). The \(x\) -component of the electric field at a point \((a, 0)\) is given by $$E_{x}(a)=\frac{k Q a}{2 L} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{3 / 2}}$$ where \(k\) is a physical constant and \(a>0\) a. Confirm that \(E_{x}(a)=\frac{k Q}{a \sqrt{a^{2}+L^{2}}}\) b. Letting \(\rho=Q / 2 L\) be the charge density on the line segment, show that if \(L \rightarrow \infty,\) then \(E_{x}(a)=2 k \rho / a\) (See the Guided Project Electric field integrals for a derivation of this and other similar integrals.)
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