Solve ∫ (from 1 to 3) of 6x-x^2 wrt x

$$\int_{ 1 }^{ 3 } 6x- { x }^{ 2 } d x$$

Evaluate

$$\frac{46}{3}\approx 15.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -1 times \frac{x^{3}}{3}.

$$3x^{2}-\frac{x^{3}}{3}$$

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.

$$3\times 3^{2}-\frac{3^{3}}{3}-\left(3\times 1^{2}-\frac{1^{3}}{3}\right)$$

Simplify.

$$\frac{46}{3}$$