Solve ∫ (from 6 to 9) of left(x-9right)left(4x-36right) wrt x

$$\int_{ 6 }^{ 9 } \left( x-9 \right) \left( 4x-36 \right) d x$$

Evaluate

$$36$$

Solution Steps

Apply the distributive property by multiplying each term of x-9 by each term of 4x-36.

$$\int _{6}^{9}4x^{2}-36x-36x+324\mathrm{d}x$$

Combine -36x and -36x to get -72x.

$$\int _{6}^{9}4x^{2}-72x+324\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x^{2}-72x+324\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int -72x\mathrm{d}x+\int 324\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x-72\int x\mathrm{d}x+\int 324\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}-72\int x\mathrm{d}x+\int 324\mathrm{d}x$$

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 -72 times \frac{x^{2}}{2}.

$$\frac{4x^{3}}{3}-36x^{2}+\int 324\mathrm{d}x$$

Find the integral of 324 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{4x^{3}}{3}-36x^{2}+324x$$

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.

$$\frac{4}{3}\times 9^{3}-36\times 9^{2}+324\times 9-\left(\frac{4}{3}\times 6^{3}-36\times 6^{2}+324\times 6\right)$$

Simplify.

$$36$$