Solve ∫ (from 0 to 2) of 3(x+1)(x+6) wrt x=

$$\int _ { 0 } ^ { 2 } 3 ( x + 1 ) ( x + 6 ) d x =$$

Evaluate

$$86$$

Solution Steps

Use the distributive property to multiply 3 by x+1.

$$\int _{0}^{2}\left(3x+3\right)\left(x+6\right)\mathrm{d}x$$

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

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

Combine 18x and 3x to get 21x.

$$\int _{0}^{2}3x^{2}+21x+18\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 3x^{2}+21x+18\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{2}\mathrm{d}x+\int 21x\mathrm{d}x+\int 18\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{2}\mathrm{d}x+21\int x\mathrm{d}x+\int 18\mathrm{d}x$$

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}.

$$x^{3}+21\int x\mathrm{d}x+\int 18\mathrm{d}x$$

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

$$x^{3}+\frac{21x^{2}}{2}+\int 18\mathrm{d}x$$

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

$$x^{3}+\frac{21x^{2}}{2}+18x$$

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.

$$2^{3}+\frac{21}{2}\times 2^{2}+18\times 2-\left(0^{3}+\frac{21}{2}\times 0^{2}+18\times 0\right)$$

Simplify.

$$86$$