Find the value of A if $\int_{0}^{\sqrt{3}} \frac{x^{2} d x}{1+x^{2}}=\sqrt{3}-\frac{\pi}{A}$ .
A)2
B)3
C)4
D)6

Question

Quizplus · Accepted Answer

The integral can be solved by recognizing it as a standard form. Let's solve the integral first:$$\int_{0}^{\sqrt{3}} \frac{x^{2}}{1+x^{2}} dx$$This can be rewritten as:$$\int_{0}^{\sqrt{3}} \left(1 - \frac{1}{1+x^{2}}\right) dx$$Splitting the integral, we get:$$\int_{0}^{\sqrt{3}} 1 dx - \int_{0}^{\sqrt{3}} \frac{1}{1+x^{2}} dx$$Solving each part:$$x \Big|_{0}^{\sqrt{3}} - \tan^{-1}(x) \Big|_{0}^{\sqrt{3}}$$Substituting the limits:$$\sqrt{3} - 0 - (\tan^{-1}(\sqrt{3}) - \tan^{-1}(0))$$$$\sqrt{3} - \frac{\pi}{3}$$Comparing this with the given equation:$$\sqrt{3} - \frac{\pi}{A} = \sqrt{3} - \frac{\pi}{3}$$Hence, $A = 3$.

Find the Value of a If ∫03x2dx1+x2=3−πA\int_{0}^{\sqrt{3}} \frac{x^{2} d x}{1+x^{2}}=\sqrt{3}-\frac{\pi}{A}∫03​​1+x2x2dx​=3​−Aπ​ A)2 B)3 C)4 D)6

Find the Value of a If $\int_{0}^{\sqrt{3}} \frac{x^{2} d x}{1+x^{2}}=\sqrt{3}-\frac{\pi}{A}$
A)2
B)3
C)4
D)6