Solve ∫ (from 0 to 1) of (u^2+6x)du

$$\int _ { 0 } ^ { 1 } ( u ^ { 2 } + 6 x ) d u$$

Evaluate

$$6x+\frac{1}{3}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int u^{2}+6x\mathrm{d}u$$

Integrate the sum term by term.

$$\int u^{2}\mathrm{d}u+\int 6x\mathrm{d}u$$

Factor out the constant in each of the terms.

$$\int u^{2}\mathrm{d}u+6\int x\mathrm{d}u$$

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}.

$$\frac{u^{3}}{3}+6\int x\mathrm{d}u$$

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

$$\frac{u^{3}}{3}+6xu$$

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.

$$\frac{1^{3}}{3}+6x\times 1-\left(\frac{0^{3}}{3}+6x\times 0\right)$$

Simplify.

$$\frac{1}{3}+6x$$

Differentiate w.r.t. x

$$6$$