Solve ∫ (from 0 to 3) of (1-u)du

$$\int _ { 0 } ^ { 3 } ( 1 - u ) d u$$

Evaluate

$$-\frac{3}{2}=-1.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 1-u\mathrm{d}u$$

Integrate the sum term by term.

$$\int 1\mathrm{d}u+\int -u\mathrm{d}u$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}u-\int u\mathrm{d}u$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}u=au.

$$u-\int u\mathrm{d}u$$

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

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

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.

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

Simplify.

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