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

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

Evaluate

$$4$$

Solution Steps

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

$$\int _{1}^{2}3x^{2}-9x-7x+21\mathrm{d}x$$

Combine -9x and -7x to get -16x.

$$\int _{1}^{2}3x^{2}-16x+21\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 3x^{2}-16x+21\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{2}\mathrm{d}x+\int -16x\mathrm{d}x+\int 21\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{2}\mathrm{d}x-16\int x\mathrm{d}x+\int 21\mathrm{d}x$$

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

$$x^{3}-16\int x\mathrm{d}x+\int 21\mathrm{d}x$$

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 -16 times \frac{x^{2}}{2}.

$$x^{3}-8x^{2}+\int 21\mathrm{d}x$$

Find the integral of 21 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{3}-8x^{2}+21x$$

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.

$$2^{3}-8\times 2^{2}+21\times 2-\left(1^{3}-8\times 1^{2}+21\times 1\right)$$

Simplify.

$$4$$