Solve ∫ (from 2 to 3) of x^2+3x^3 wrt x

$$\int _ { 2 } ^ { 3 } x ^ { 2 } + 3 x ^ { 3 } d x$$

Evaluate

$$\frac{661}{12}\approx 55.083333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{x^{3}}{3}+3\int x^{3}\mathrm{d}x$$

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

$$\frac{x^{3}}{3}+\frac{3x^{4}}{4}$$

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.

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

Simplify.

$$\frac{661}{12}$$