JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(f:(0, \infty) \rightarrow R\) and \(F(x)=\int_0^x t f(t) d t\). If \(F\left(x^2\right)=\) \(x^4+x^5\), then \(\sum_{r=1}^{12} f\left(r^2\right)\) is equal to :
- A \(345\)
- B \(245\)
- C \(219\)
- D \(456\)
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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