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Chapter 11: Problem 40

Use a symbolic integration utility to find the indefinite integral. $$ \int \sqrt{x}(x+1) d x $$

Short Answer

Expert verified
The indefinite integral of \(\sqrt{x}(x+1) dx\) is \(\frac{x^3}{3} + x + C\)

Step by step solution

01

Expand the function

Expand the function inside the integral, \(\sqrt{x}(x+1) = x\sqrt{x} + \sqrt{x}\). So our integral becomes \(\int (x\sqrt{x} + \sqrt{x}) dx\).
02

Separate the Integral

Based on the properties of integrals, we can separate this integral into two parts: \(\int x\sqrt{x} dx + \int \sqrt{x} dx\)
03

Change of Variable

Take \(u = x^{3/2}\). Then, \(\frac{du}{dx} = \frac{3}{2} x^{1/2}\) or \(dx = \frac{2du}{3x^{1/2}}\). Substitute back to the first integral, we have \(\int \frac{2}{3}u du\). Take \(v = x^{1/2}\). Then, \( \frac{dv}{dx} = \frac{1}{2} x^{-1/2}\) or \(dx = 2 dv x^{1/2}\). Substitute back to the second integral, we have \(\int 2v du \).
04

Integrate

Now we can integrate the functions. For the first integral, the integral of \(u\) is \(\frac{u^2}{2} = \frac{x^3}{2}\). For the second integral, the integral of \(v\) is \(\frac{v^2}{2} = \frac{x}{2}\). So, the indefinite integral will be: \(\frac{2}{3} \cdot \frac{x^3}{2} + 2 \cdot \frac{x}{2} + C\)

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Most popular questions from this chapter

Use the Midpoint Rule with \(n=4\) to approximate the area of the region bounded by the graph of \(f\) and the \(x\) -axis over the interval. Compare your result with the exact area. Sketch the region. $$ f(x)=2 x^{2} $$

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ y=x^{3}-2 x+1, y=-2 x, x=1 $$

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ \begin{aligned} &y=x e^{-x^{2}}, y=0, x=0, x=1\\\ &\begin{gathered} 51 / 3 \end{gathered} \end{aligned} $$

Use the value \(\int_{0}^{2} x^{3} d x=4\) to evaluate each definite integral. Explain your reasoning. (a) \(\int_{-2}^{0} x^{3} d x\) (b) \(\int_{-2}^{2} x^{3} d x\) (c) \(\int_{0}^{2} 3 x^{3} d x\)

Use the Midpoint Rule with \(n=4\) to approximate the area of the region. Compare your result with the exact area obtained with a definite integral. $$ f(y)=2 y, \quad[0,2] $$
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