/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 35

Evaluate the following integrals or state that they diverge. $$\int_{0}^{8} \frac{d x}{\sqrt[3]{x}}$$

Short Answer

Expert verified
Answer: The value of the definite integral is 6.

Step by step solution

01

Rewrite the function

Rewrite the function as $$\frac{1}{\sqrt[3]{x}} = x^{-\frac{1}{3}}.$$ This makes it easier to find the antiderivative.
02

Find the antiderivative

Apply the power rule for integration: $$\int x^n dx = \frac{x^{n+1}}{n+1} + C, \text{ for } n \neq -1.$$ In our case, \(n = -\frac{1}{3}\). So the antiderivative, F(x), is: $$F(x) = \frac{x^{(-\frac{1}{3} + 1)}}{(-\frac{1}{3} + 1)} + C = \frac{x^{\frac{2}{3}}}{\frac{2}{3}} + C = \frac{3}{2}x^{\frac{2}{3}} + C.$$
03

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if a function \(f\) is continuous on the interval \([a, b]\), and \(F\) is its antiderivative, then $$\int_a^b f(x)dx = F(b) - F(a).$$ In our case, \(a = 0\) and \(b = 8\). Applying the theorem, we get: $$\int_0^8 \frac{1}{\sqrt[3]{x}} dx = \left[\frac{3}{2}x^{\frac{2}{3}}\right]_0^8 = \frac{3}{2}(8^{\frac{2}{3}}) - \frac{3}{2}(0^{\frac{2}{3}}).$$
04

Evaluate the integral

Now, let's evaluate the definite integral: $$\frac{3}{2}(8^{\frac{2}{3}}) - \frac{3}{2}(0^{\frac{2}{3}}) = \frac{3}{2}(4) - 0 = 6.$$ Hence, the value of the definite integral is \(\int_0^8 \frac{1}{\sqrt[3]{x}} dx = 6\). The integral converges and its value is 6.

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Most popular questions from this chapter

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