Solve ∫ (from 1 to 3) of -x^2+2x+3 wrt x

$$\int _ { 1 } ^ { 3 } - x ^ { 2 } + 2 x + 3 d x$$

Evaluate

$$\frac{16}{3}\approx 5.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 -1 times \frac{x^{3}}{3}.

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

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}.

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

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$\frac{16}{3}$$