Solve ∫ (from 0 to 1) of (x+3)(3x-6) wrt x=

$$\int _ { 0 } ^ { 1 } ( x + 3 ) ( 3 x - 6 ) d x =$$

Evaluate

$$-\frac{31}{2}=-15.5$$

Solution Steps

Apply the distributive property by multiplying each term of x+3 by each term of 3x-6.

$$\int _{0}^{1}3x^{2}-6x+9x-18\mathrm{d}x$$

Combine -6x and 9x to get 3x.

$$\int _{0}^{1}3x^{2}+3x-18\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 3x^{2}+3x-18\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{2}\mathrm{d}x+\int 3x\mathrm{d}x+\int -18\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{2}\mathrm{d}x+3\int x\mathrm{d}x+\int -18\mathrm{d}x$$

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}.

$$x^{3}+3\int x\mathrm{d}x+\int -18\mathrm{d}x$$

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

$$x^{3}+\frac{3x^{2}}{2}+\int -18\mathrm{d}x$$

Find the integral of -18 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{3}+\frac{3x^{2}}{2}-18x$$

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.

$$1^{3}+\frac{3}{2}\times 1^{2}-18-\left(0^{3}+\frac{3}{2}\times 0^{2}-18\times 0\right)$$

Simplify.

$$-\frac{31}{2}$$