Solve ∫ (from 2 to 3) of 8x^3+3x^2+6x

$$\int _ { 2 } ^ { 3 } 8 x ^ { 3 } + 3 x ^ { 2 } + 6 x$$

Evaluate

$$164$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 8x^{3}+3x^{2}+6x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x^{3}\mathrm{d}x+\int 3x^{2}\mathrm{d}x+\int 6x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{3}\mathrm{d}x+3\int x^{2}\mathrm{d}x+6\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 8 times \frac{x^{4}}{4}.

$$2x^{4}+3\int x^{2}\mathrm{d}x+6\int x\mathrm{d}x$$

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}.

$$2x^{4}+x^{3}+6\int x\mathrm{d}x$$

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

$$2x^{4}+x^{3}+3x^{2}$$

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.

$$2\times 3^{4}+3^{3}+3\times 3^{2}-\left(2\times 2^{4}+2^{3}+3\times 2^{2}\right)$$

Simplify.

$$164$$