Solve ∫ (from - 2 to 0) of (7x^2+7x) wrt x

$$\int _ { - 2 } ^ { 0 } ( 7 x ^ { 2 } + 7 x ) d x$$

Evaluate

$$\frac{14}{3}\approx 4.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 7x^{2}+7x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 7x^{2}\mathrm{d}x+\int 7x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$7\int x^{2}\mathrm{d}x+7\int x\mathrm{d}x$$

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}. Multiply 7 times \frac{x^{3}}{3}.

$$\frac{7x^{3}}{3}+7\int x\mathrm{d}x$$

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 7 times \frac{x^{2}}{2}.

$$\frac{7x^{3}}{3}+\frac{7x^{2}}{2}$$

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.

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

Simplify.

$$\frac{14}{3}$$