Solve ∫ (from 0 to 2) of 2x*3/sqrt{3*sqrt{2 wrt x

$$\int_{ 0 }^{ 2 } \frac{ 2x \times 3 }{ \sqrt{ 3 } \times \sqrt{ 2 } } d x$$

Evaluate

$$2\sqrt{6}\approx 4.898979486$$

Solution Steps

Multiply 2 and 3 to get 6.

$$\int _{0}^{2}\frac{6x}{\sqrt{3}\sqrt{2}}\mathrm{d}x$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

Rationalize the denominator of \frac{6x}{\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.

$$\int _{0}^{2}\frac{6x\sqrt{6}}{\left(\sqrt{6}\right)^{2}}\mathrm{d}x$$

The square of \sqrt{6} is 6.

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

Cancel out 6 and 6.

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

Evaluate the indefinite integral first.

$$\int x\sqrt{6}\mathrm{d}x$$

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

$$\sqrt{6}\int x\mathrm{d}x$$

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}.

$$\sqrt{6}\times \frac{x^{2}}{2}$$

Simplify.

$$\frac{\sqrt{6}x^{2}}{2}$$

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.

$$\frac{1}{2}\times 6^{\frac{1}{2}}\times 2^{2}-\frac{1}{2}\times 6^{\frac{1}{2}}\times 0^{2}$$

Simplify.

$$2\sqrt{6}$$