JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(f\) be a continuous function satisfying \(\int \limits_0^{t^2}\left( f ( x )+ x ^2\right) dx =\frac{4}{3} t ^3, \forall t > 0 . \quad\) Then \(f \left(\frac{\pi^2}{4}\right)\) equal to :
- A \(\pi\left(1-\frac{\pi^3}{16}\right)\)
- B \(-\pi^2\left(1+\frac{\pi^2}{16}\right)\)
- C \(-\pi\left(1+\frac{\pi^3}{16}\right)\)
- D \(\pi^2\left(1-\frac{\pi^2}{16}\right)\)
Answer & Solution
Correct Answer
(A) \(\pi\left(1-\frac{\pi^3}{16}\right)\)
Step-by-step Solution
Detailed explanation
\(\int \limits_0^{t^2}\left( f ( x )+ x ^2\right) d x=\frac{4}{3} t ^3, \forall t > 0\) \(\left( f \left( t ^2\right)+ t ^4\right)=2 t\) \(f \left( t ^2\right)=2 t - t ^4\) \(t =\frac{\pi}{2} \Rightarrow f \left(\frac{\pi^2}{4}\right)=\frac{2 \pi}{2}-\frac{\pi^4}{16}\)…
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