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JEE Mains · Maths · STD 12 - 7.2 definite integral

माना \(\mathrm{f}:(0, \infty) \rightarrow \mathrm{R}\) तथा \(\mathrm{F}(\mathrm{x})=\int_0^x \mathrm{tf}(\mathrm{t}) \mathrm{dt}\) है। यदि \(\mathrm{F}\left(\mathrm{x}^2\right)=\mathrm{x}^4+\mathrm{x}^5\) है तो \(\sum_{\mathrm{r}=1}^{12} \mathrm{f}\left(\mathrm{r}^2\right)\) बराबर है ........

  1. A \(345\)
  2. B \(245\)
  3. C \(219\)
  4. D \(456\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(219\)

Step-by-step Solution

Detailed explanation

\( F(x)=\int_0^x t \cdot f(t) d t \) \( \mathrm{~F}^1(\mathrm{x})=\mathrm{xf}(\mathrm{x}) \) \( \text { Given } \) \( F\left(x^2\right)=x^4+x^5, \quad \text { let } x^2=t \) \( F(t)=t^2+t^{5 / 2} \) \( F^{\prime}(t)=2 t+5 / 2 t^{3 / 2} \) \( t \cdot f(t)=2 t+5 / 2 t^{3 / 2} \)…
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