Solve 1/sqrt{3}[frac{x^{2}}{2}]_{0}^sqrt{3}*(frac{xsqrt{4-x^2}}{2}+4/2sin^-1x/2]_{sqrt{3}}^2

$$\frac { 1 } { \sqrt { 3 } } [ \frac { x ^ { 2 } } { 2 } ] _ { 0 } ^ { \sqrt { 3 } } \cdot ( \frac { x \sqrt { 4 - x ^ { 2 } } } { 2 } + \frac { 4 } { 2 } \sin ^ { - 1 } \frac { x } { 2 } ] _ { \sqrt { 3 } } ^ { 2 }$$

Evaluate

$$\frac{\sqrt{3}x^{2}}{6}$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

$$\frac{\sqrt{3}}{3}\times \frac{x^{2}}{2}$$

Multiply \frac{\sqrt{3}}{3} times \frac{x^{2}}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\sqrt{3}x^{2}}{3\times 2}$$

Multiply 3 and 2 to get 6.

$$\frac{\sqrt{3}x^{2}}{6}$$

Differentiate w.r.t. x

$$\frac{\sqrt{3}x}{3}$$