Solve ∫ (from 0 to 3) of (-26+6x)(-6+x) wrt x

$$\int_{ 0 }^{ 3 } (-26+6x)(-6+x) d x$$

Evaluate

$$243$$

Solution Steps

Apply the distributive property by multiplying each term of -26+6x by each term of -6+x.

$$\int _{0}^{3}156-26x-36x+6x^{2}\mathrm{d}x$$

Combine -26x and -36x to get -62x.

$$\int _{0}^{3}156-62x+6x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 156-62x+6x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 156\mathrm{d}x+\int -62x\mathrm{d}x+\int 6x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 156\mathrm{d}x-62\int x\mathrm{d}x+6\int x^{2}\mathrm{d}x$$

Find the integral of 156 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$156x-62\int x\mathrm{d}x+6\int x^{2}\mathrm{d}x$$

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 -62 times \frac{x^{2}}{2}.

$$156x-31x^{2}+6\int x^{2}\mathrm{d}x$$

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

$$156x-31x^{2}+2x^{3}$$

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.

$$156\times 3-31\times 3^{2}+2\times 3^{3}-\left(156\times 0-31\times 0^{2}+2\times 0^{3}\right)$$

Simplify.

$$243$$