Solve ∫ (from - 2 to 5) of x^2+2x wrt x

$$\int _ { - 2 } ^ { 5 } x ^ { 2 } + 2 x d x$$

Evaluate

$$\frac{196}{3}\approx 65.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{x^{3}}{3}+2\int x\mathrm{d}x$$

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

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

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.

$$\frac{5^{3}}{3}+5^{2}-\left(\frac{\left(-2\right)^{3}}{3}+\left(-2\right)^{2}\right)$$

Simplify.

$$\frac{196}{3}$$