Solve ∫ (from 0 to 6) of 2x-5-x^2+4x

$$\int _ { 0 } ^ { 6 } 2 x - 5 - x ^ { 2 } + 4 x$$

Evaluate

$$6$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

Find the integral of -5 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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

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

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

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

Simplify.

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

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.

$$3\times 6^{2}-5\times 6-\frac{6^{3}}{3}-\left(3\times 0^{2}-5\times 0-\frac{0^{3}}{3}\right)$$

Simplify.

$$6$$