Evaluate the integral: \int^1_0 \sqrt x(1- x^2) dx


Evaluate the integral: {eq}\int^1_0 \sqrt x(1- x^2) dx {/eq}

Definite Integral:

If given two functions, we need to distribute or expand. Solving the definite integral of a function at a given interval requires knowledge of the fundamental theorem of calculus. The fundamental theorem of calculus is defined by {eq}\int_{a}^{b} f(x) dx = F(a)-F(b) {/eq}.

Answer and Explanation: 1

Given the function, we are asked to evaluate the integral. First, we need to expand the function, we have

{eq}\begin{align} \int^1_0 \sqrt x(1- x^2) dx &= \int^1_0 \sqrt{x}-x^2 \sqrt{x} dx \\ &= \int^1_0 \sqrt{x}- x^2(x^{\frac{1}{2}}) dx \\ &= \int^1_0 \sqrt{x}- x^{\frac{5}{2}} dx \\ &= \int^1_0 \sqrt{x} dx - \int^1_0 x^{\frac{5}{2}} dx \\ &= \int^1_0 x^{\frac{1}{2}} dx - \int^1_0 x^{\frac{5}{2}} dx \\ &= \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{\frac{7}{2}}}{\frac{7}{2}} \big|^1_0 \\ &= \frac{2}{3}x^{\frac{3}{2}}- \frac{2}{7}x^{\frac{7}{2}}\big|^1_0 \\ &= \frac{2}{3}(1)^{\frac{3}{2}}- \frac{2}{7}(1)^{\frac{7}{2}} - \frac{2}{3}(0)^{\frac{3}{2}}- \frac{2}{7}(0)^{\frac{7}{2}} \\ &= \frac{8}{21} - 0 \\ &= \frac{8}{21} \end{align} {/eq}

Learn more about this topic:

Evaluating Definite Integrals Using the Fundamental Theorem


Chapter 16 / Lesson 2

In calculus, the fundamental theorem is an essential tool that helps explain the relationship between integration and differentiation. Learn about evaluating definite integrals using the fundamental theorem, and work examples to gain understanding.

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