Solve ∫ (from 0 to 7) of (u^2-4u)du

$$\int _ { 0 } ^ { 7 } ( u ^ { 2 } - 4 u ) d u$$

Evaluate

$$\frac{49}{3}\approx 16.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int u^{2}-4u\mathrm{d}u$$

Integrate the sum term by term.

$$\int u^{2}\mathrm{d}u+\int -4u\mathrm{d}u$$

Factor out the constant in each of the terms.

$$\int u^{2}\mathrm{d}u-4\int u\mathrm{d}u$$

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

$$\frac{u^{3}}{3}-4\int u\mathrm{d}u$$

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}. Multiply -4 times \frac{u^{2}}{2}.

$$\frac{u^{3}}{3}-2u^{2}$$

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.

$$\frac{7^{3}}{3}-2\times 7^{2}-\left(\frac{0^{3}}{3}-2\times 0^{2}\right)$$

Simplify.

$$\frac{49}{3}$$