Solve ∫ (from 0 to 3) of (135+95x)(6+x) wrt x

$$\int_{ 0 }^{ 3 } (135+95x)(6+x) d x$$

Evaluate

$$\frac{12915}{2}=6457.5$$

Solution Steps

Apply the distributive property by multiplying each term of 135+95x by each term of 6+x.

$$\int _{0}^{3}810+135x+570x+95x^{2}\mathrm{d}x$$

Combine 135x and 570x to get 705x.

$$\int _{0}^{3}810+705x+95x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 810+705x+95x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 810\mathrm{d}x+\int 705x\mathrm{d}x+\int 95x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 810\mathrm{d}x+705\int x\mathrm{d}x+95\int x^{2}\mathrm{d}x$$

Find the integral of 810 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$810x+705\int x\mathrm{d}x+95\int x^{2}\mathrm{d}x$$

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

$$810x+\frac{705x^{2}}{2}+95\int x^{2}\mathrm{d}x$$

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

$$810x+\frac{705x^{2}}{2}+\frac{95x^{3}}{3}$$

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.

$$810\times 3+\frac{705}{2}\times 3^{2}+\frac{95}{3}\times 3^{3}-\left(810\times 0+\frac{705}{2}\times 0^{2}+\frac{95}{3}\times 0^{3}\right)$$

Simplify.

$$\frac{12915}{2}$$