Solve ∫ (from 0 to 5) of -x^2+x+6-x-3

$$\int _ { 0 } ^ { 5 } - x ^ { 2 } + x + 6 - x - 3$$

Evaluate

$$-\frac{80}{3}\approx -26.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -x^{2}+x+6-x-3\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 6\mathrm{d}x+\int -x\mathrm{d}x+\int -3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 6\mathrm{d}x-\int x\mathrm{d}x+\int -3\mathrm{d}x$$

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}.

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

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}.

$$-\frac{x^{3}}{3}+\frac{x^{2}}{2}+\int 6\mathrm{d}x-\int x\mathrm{d}x+\int -3\mathrm{d}x$$

Find the integral of 6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{x^{3}}{3}+\frac{x^{2}}{2}+6x-\int x\mathrm{d}x+\int -3\mathrm{d}x$$

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 -1 times \frac{x^{2}}{2}.

$$-\frac{x^{3}}{3}+\frac{x^{2}}{2}+6x-\frac{x^{2}}{2}+\int -3\mathrm{d}x$$

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

$$-\frac{x^{3}}{3}+\frac{x^{2}}{2}+6x-\frac{x^{2}}{2}-3x$$

Simplify.

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

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.

$$-\frac{5^{3}}{3}+3\times 5-\left(-\frac{0^{3}}{3}+3\times 0\right)$$

Simplify.

$$-\frac{80}{3}$$