/*! 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 31 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 31

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{1}(x+\sqrt{x}) d x$$

Short Answer

Expert verified
Based on the given step by step solution, the integral of the function (x + 鈭歺) from 0 to 1 is 7/6.

Step by step solution

01

Find the antiderivatives of x and 鈭歺

To find the antiderivative of x, we can use the power rule, which states that the antiderivative of x^n (where n 鈮� -1) is (x^(n+1))/(n+1) + C. Applying this rule, we have: Antiderivative of x = (x^(1+1))/(1+1) + C = x^2 / 2 + C To find the antiderivative of 鈭歺, we first need to rewrite the term using fractional exponents: 鈭歺 = x^(1/2). Now we can apply the power rule: Antiderivative of x^(1/2) = (x^((1/2)+1))/((1/2)+1) + D = x^(3/2) / (3/2) + D = (2/3)x^(3/2) + D.
02

Combine the antiderivatives

Antiderivative of (x + 鈭歺) = x^2 / 2 + (2/3)x^(3/2) + (C + D) = x^2 / 2 + (2/3)x^(3/2) + K, where K is a constant of integration.
03

Apply the Fundamental Theorem of Calculus

Now we will use the antiderivative found in Step 2 to evaluate the integral from 0 to 1: $$\int_{0}^{1}(x+\sqrt{x}) d x = \left[\frac{x^2}{2}+\frac{2}{3}x^{\frac{3}{2}}\right]_{0}^{1}$$ $$\Rightarrow \left(\frac{1^2}{2}+\frac{2}{3}1^{\frac{3}{2}}\right)-\left(\frac{0^2}{2}+\frac{2}{3}0^{\frac{3}{2}}\right) = \frac{1}{2}+\frac{2}{3}$$ $$\Rightarrow \frac{3+4}{6} = \frac{7}{6}$$ Thus, the evaluated integral is 7/6.

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Most popular questions from this chapter

Multiple substitutions Use two or more substitutions to find the following integrals. $$\int x \sin ^{4}\left(x^{2}\right) \cos \left(x^{2}\right) d x$$ (Hint: Begin with \(u=x^{2}\), then use \(v=\sin u .)\)

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