Solve ∫ (from 1 / 3 to frac { 1) of {2}}x^2+x^3

$$\int _ { 1 / 3 } ^ { \frac { 1 } { 2 } } x ^ { 2 } + x ^ { 3 }$$

Evaluate

$$\frac{217}{5184}\approx 0.041859568$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}+x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

$$\frac{x^{3}}{3}+\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{3}}{3}+\frac{x^{4}}{4}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{\left(\frac{1}{2}\right)^{3}}{3}+\frac{\left(\frac{1}{2}\right)^{4}}{4}-\left(\frac{\left(\frac{1}{3}\right)^{3}}{3}+\frac{\left(\frac{1}{3}\right)^{4}}{4}\right)$$

Simplify.

$$\frac{217}{5184}$$