Solve ∫ (from 1 to 3) of 1+5x-x^2-4x-x^2 wrt x

$$\int _ { 1 } ^ { 3 } 1 + 5 x - x ^ { 2 } - 4 x - x ^ { 2 } d x$$

Evaluate

$$-\frac{34}{3}\approx -11.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 1+5x-x^{2}-4x-x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 1\mathrm{d}x+\int 5x\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int -4x\mathrm{d}x+\int -x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}x+5\int x\mathrm{d}x-\int x^{2}\mathrm{d}x-4\int x\mathrm{d}x-\int x^{2}\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x+5\int x\mathrm{d}x-\int x^{2}\mathrm{d}x-4\int x\mathrm{d}x-\int x^{2}\mathrm{d}x$$

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

$$x+\frac{5x^{2}}{2}-\int x^{2}\mathrm{d}x-4\int x\mathrm{d}x-\int x^{2}\mathrm{d}x$$

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}.

$$x+\frac{5x^{2}}{2}-\frac{x^{3}}{3}-4\int x\mathrm{d}x-\int x^{2}\mathrm{d}x$$

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 -4 times \frac{x^{2}}{2}.

$$x+\frac{5x^{2}}{2}-\frac{x^{3}}{3}-2x^{2}-\int x^{2}\mathrm{d}x$$

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}.

$$x+\frac{5x^{2}}{2}-\frac{x^{3}}{3}-2x^{2}-\frac{x^{3}}{3}$$

Simplify.

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

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.

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

Simplify.

$$-\frac{34}{3}$$