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Chapter 7: Problem 42

Use the approaches discussed in this section to evaluate the following integrals. $$\int_{4}^{9} \frac{d x}{1-\sqrt{x}}$$

Short Answer

Expert verified
Question: Evaluate the integral $\int_{4}^{9} \frac{d x}{1-\sqrt{x}}$. Answer: 5

Step by step solution

01

Choose a substitution variable

Let's choose the substitution variable u such that: $$u = 1 - \sqrt{x}$$
02

Solve for x and dx

To make the substitution, we need to find expressions for x and dx in terms of u. Solve for x: $$x = (1 - u)^2$$ Now find the derivative of x with respect to u: $$\frac{dx}{du} = -2(1-u)$$ Now, find dx by multiplying by du: $$dx = -2(1-u) du$$
03

Substitute and integrate

Replace x and dx in the integral with the expressions we found in terms of u: $$\int_{4}^{9} \frac{d x}{1-\sqrt{x}} = \int_{u(4)}^{u(9)} \frac{-2(1-u) du}{u}$$ Now we need to find the new bounds of the integral. Since u = 1 - 鈭歺, find u(4) and u(9): $$u(4) = 1 - \sqrt{4} = -1$$ $$u(9) = 1 - \sqrt{9} = -2$$ Now plug in those values into the integral: $$\int_{-1}^{-2} \frac{-2(1-u) du}{u}$$ Now we can integrate the simplified expression: $$\int_{-1}^{-2}(-2 + 2u) du$$
04

Evaluate the antiderivative

Find the antiderivative: $$\int(-2 + 2u)du = -2u + u^2 + C$$ Apply the Fundamental Theorem of Calculus over the integral bounds: $$\left(-2(-2) + (-2)^2\right) - \left(-2(-1) + (-1)^2\right) = 4 + 4 - (2 + 1) = 5$$ The value of the integral is 5.

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