Solve ∫ (from 1 to 3) of x^3+x

$$\int _ { 1 } ^ { 3 } x ^ { 3 } + x$$

Evaluate

$$24$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

$$\frac{x^{4}}{4}+\int x\mathrm{d}x$$

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

$$\frac{x^{4}}{4}+\frac{x^{2}}{2}$$

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.

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

Simplify.

$$24$$