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

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

Evaluate

$$\frac{9}{2}=4.5$$

Solution Steps

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

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

Combine 8x and -x to get 7x.

$$\int _{-1}^{2}2x^{2}+7x-4\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2x^{2}+7x-4\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int 7x\mathrm{d}x+\int -4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x+7\int x\mathrm{d}x+\int -4\mathrm{d}x$$

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

$$\frac{2x^{3}}{3}+7\int x\mathrm{d}x+\int -4\mathrm{d}x$$

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

$$\frac{2x^{3}}{3}+\frac{7x^{2}}{2}+\int -4\mathrm{d}x$$

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{3}}{3}+\frac{7x^{2}}{2}-4x$$

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.

$$\frac{2}{3}\times 2^{3}+\frac{7}{2}\times 2^{2}-4\times 2-\left(\frac{2}{3}\left(-1\right)^{3}+\frac{7}{2}\left(-1\right)^{2}-4\left(-1\right)\right)$$

Simplify.

$$\frac{9}{2}$$