Solve ∫ (from 1 to 2) of (3x-7)(2-4x) wrt x

$$\int _ { 1 } ^ { 2 } ( 3 x - 7 ) ( 2 - 4 x ) d x$$

Evaluate

$$9$$

Solution Steps

Apply the distributive property by multiplying each term of 3x-7 by each term of 2-4x.

$$\int _{1}^{2}6x-12x^{2}-14+28x\mathrm{d}x$$

Combine 6x and 28x to get 34x.

$$\int _{1}^{2}34x-12x^{2}-14\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 34x-12x^{2}-14\mathrm{d}x$$

Integrate the sum term by term.

$$\int 34x\mathrm{d}x+\int -12x^{2}\mathrm{d}x+\int -14\mathrm{d}x$$

Factor out the constant in each of the terms.

$$34\int x\mathrm{d}x-12\int x^{2}\mathrm{d}x+\int -14\mathrm{d}x$$

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

$$17x^{2}-12\int x^{2}\mathrm{d}x+\int -14\mathrm{d}x$$

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 -12 times \frac{x^{3}}{3}.

$$17x^{2}-4x^{3}+\int -14\mathrm{d}x$$

Find the integral of -14 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$17x^{2}-4x^{3}-14x$$

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.

$$17\times 2^{2}-4\times 2^{3}-14\times 2-\left(17\times 1^{2}-4\times 1^{3}-14\right)$$

Simplify.

$$9$$