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Chapter 5: Problem 87

Evaluate the following definite integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{4} \frac{x-2}{\sqrt{x}} d x$$

Short Answer

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Question: Evaluate the definite integral 鈭�(x-2)/鈭歺 dx from 1 to 4. Answer: -7

Step by step solution

01

Simplify the integrand

Rewrite the fraction as two separate fractions and simplify the expressions: $$\int_{1}^{4} \frac{x-2}{\sqrt{x}} d x = \int_{1}^{4} \left(\frac{x}{\sqrt{x}} - \frac{2}{\sqrt{x}}\right) d x = \int_{1}^{4} (x^{\frac{1}{2} - 1} - 2x^{-\frac{1}{2}}) dx$$
02

Find the antiderivative

Use the power rule for integration to find the antiderivative: $$\int_{1}^{4} (x^{\frac{1}{2}-1} - 2x^{-\frac{1}{2}}) d x = \left[x^{\frac{1}{2}} - \frac{4x^{\frac{1}{2}}}{\frac{1}{2}}\right] + C$$ Simplify the expression: $$\int_{1}^{4} (x^{\frac{1}{2}-1} - 2x^{-\frac{1}{2}}) d x = \left[x^{\frac{1}{2}} - 8x^{\frac{1}{2}}\right] + C$$
03

Evaluate the definite integral using the Fundamental Theorem of Calculus

Plug in the limits of integration and subtract the values: $$\int_{1}^{4} (x^{\frac{1}{2}-1} - 2x^{-\frac{1}{2}}) d x = \left(4^{\frac{1}{2}} - 8\cdot4^{\frac{1}{2}}\right) - \left(1^{\frac{1}{2}} - 8\cdot1^{\frac{1}{2}}\right) = (2-16) - (1-8) = (-14)+7$$ So the result is: $$\int_{1}^{4} \frac{x-2}{\sqrt{x}} d x = -7$$

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