Solve ∫ (from 0 to 1) of (x^2+6x) wrt x

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

Evaluate

$$\frac{10}{3}\approx 3.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}. Multiply 6 times \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}+3x^{2}$$

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.

$$\frac{1^{3}}{3}+3\times 1^{2}-\left(\frac{0^{3}}{3}+3\times 0^{2}\right)$$

Simplify.

$$\frac{10}{3}$$