Solve ∫ (from - 1 to 3) of (u^3+u)du

$$\int _ { - 1 } ^ { 3 } ( u ^ { 3 } + u ) d u$$

Evaluate

$$24$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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

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

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

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

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.

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

Simplify.

$$24$$