Solve ∫ (from 0 to 3) of (13.5+9.5x)(6+x) wrt x

$$\int_{ 0 }^{ 3 } (13.5+9.5x)(6+x) d x$$

Evaluate

$$645.75$$

Solution Steps

Apply the distributive property by multiplying each term of 13.5+9.5x by each term of 6+x.

$$\int _{0}^{3}81+13.5x+57x+9.5x^{2}\mathrm{d}x$$

Combine 13.5x and 57x to get 70.5x.

$$\int _{0}^{3}81+70.5x+9.5x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 81+\frac{141x}{2}+\frac{19x^{2}}{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 81\mathrm{d}x+\int \frac{141x}{2}\mathrm{d}x+\int \frac{19x^{2}}{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 81\mathrm{d}x+\frac{141\int x\mathrm{d}x}{2}+\frac{19\int x^{2}\mathrm{d}x}{2}$$

Find the integral of 81 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$81x+\frac{141\int x\mathrm{d}x}{2}+\frac{19\int x^{2}\mathrm{d}x}{2}$$

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

$$81x+\frac{141x^{2}}{4}+\frac{19\int x^{2}\mathrm{d}x}{2}$$

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

$$81x+\frac{141x^{2}}{4}+\frac{19x^{3}}{6}$$

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.

$$81\times 3+\frac{141}{4}\times 3^{2}+\frac{19}{6}\times 3^{3}-\left(81\times 0+\frac{141}{4}\times 0^{2}+\frac{19}{6}\times 0^{3}\right)$$

Simplify.

$$\frac{2583}{4}$$