Question

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer. (Hint: The integrands that involve

absolute values will have to be considered piecewise.)

${\int }_0^1(x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.})dx$

Short answer

${\int }_0^1(x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.})dx=-\frac{3}{20}$.

Step-by-step solution

Step 1. Given information.

A definite integral is given as${\int }_0^1(x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.})dx$.

Step 2. Using the Fundamental theorem of Calculus.

We get

$$\begin{gathered} {\int }_0^1(x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.})dx \\ ={\int }_0^1{x}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}dx-{\int }_0^1{x}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}dx \\ ={\left[\frac{x^{{\displaystyle \frac{2}{3}}+1}}{{\displaystyle \frac{2}{3}}+1}\right]}_0^1-{\left[\frac{x^{{\displaystyle \frac{1}{3}}+1}}{{\displaystyle \frac{1}{3}}+1}\right]}_0^1 \\ =\left[\frac{1^{{\displaystyle \frac{5}{3}}}}{{\displaystyle \frac{5}{3}}}\right]-\left[\frac{1^{{\displaystyle \frac{4}{3}}}}{{\displaystyle \frac{4}{3}}}\right] \\ =\frac{3}{5}-\frac{3}{4} \\ =-\frac{3}{20} \end{gathered}$$

The exact value of the given definite integral is$-\frac{3}{20}$.

Step 3. The graph to verify the answer is

The solution is area under graph which is

$$\begin{gathered} a\approx -0.15 \\ \approx -\frac{3}{20} \end{gathered}$$