Solve ∫ (from 0 to 6) of (sqrt{6}x-x)^2 wrt x

$$\int _ { 0 } ^ { 6 } ( \sqrt { 6 } x - x ) ^ { 2 } d x$$

Evaluate

$$504-144\sqrt{6}\approx 151.273477039$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \left(\sqrt{6}x-x\right)^{2}\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$\left(\sqrt{6}-1\right)^{2}\int x^{2}\mathrm{d}x$$

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}.

$$\left(\sqrt{6}-1\right)^{2}\times \frac{x^{3}}{3}$$

Simplify.

$$\frac{\left(7-2\sqrt{6}\right)x^{3}}{3}$$

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.

$$\frac{1}{3}\left(7-2\times 6^{\frac{1}{2}}\right)\times 6^{3}-\frac{1}{3}\left(7-2\times 6^{\frac{1}{2}}\right)\times 0^{3}$$

Simplify.

$$504-144\sqrt{6}$$