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

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

Evaluate

$$\frac{236}{3}\approx 78.666666667$$

Solution Steps

Apply the distributive property by multiplying each term of 2x-1 by each term of 4x+5.

$$\int _{-1}^{3}8x^{2}+10x-4x-5\mathrm{d}x$$

Combine 10x and -4x to get 6x.

$$\int _{-1}^{3}8x^{2}+6x-5\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 8x^{2}+6x-5\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x^{2}\mathrm{d}x+\int 6x\mathrm{d}x+\int -5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{2}\mathrm{d}x+6\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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 8 times \frac{x^{3}}{3}.

$$\frac{8x^{3}}{3}+6\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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

$$\frac{8x^{3}}{3}+3x^{2}+\int -5\mathrm{d}x$$

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

$$\frac{8x^{3}}{3}+3x^{2}-5x$$

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.

$$\frac{8}{3}\times 3^{3}+3\times 3^{2}-5\times 3-\left(\frac{8}{3}\left(-1\right)^{3}+3\left(-1\right)^{2}-5\left(-1\right)\right)$$

Simplify.

$$\frac{236}{3}$$