Solve ∫ (from 0 to 6) of (1x)(1x) wrt x

$$\int_{ 0 }^{ 6 } (1x)(1x) d x$$

Evaluate

$$72$$

Solution Steps

Multiply x and x to get x^{2}.

$$\int _{0}^{6}1x^{2}\times 1\mathrm{d}x$$

Multiply 1 and 1 to get 1.

$$\int _{0}^{6}x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

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}.

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

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.

$$\frac{6^{3}}{3}-\frac{0^{3}}{3}$$

Simplify.

$$72$$