JEE Mains · Maths · STD 12 - 7.2 definite integral
If \([t]\) denotes the greatest integer \(\leq 1\), then the value of \(\frac{3(e-1)^2}{e} \int \limits_1^2 x^2 e^{[x]+\left[x^3\right]} d x\) is :
- A \(e^9-e\)
- B \(e ^8- e\)
- C \(e^7-1\)
- D \(e^8-1\)
Answer & Solution
Correct Answer
(B) \(e ^8- e\)
Step-by-step Solution
Detailed explanation
\(\int \limits_1^2 x ^2 e ^{\left[ x ^3\right]+1} dx\) \(x ^3= t\) \(3 x ^2 dx = dt\) \(=\frac{ e }{3} \int \limits_1^8 e ^{[t]} dt\) \(=\frac{ e }{3}\left\{\int \limits_1^2 e dt +\int_2^3 e ^2 dt +\ldots \ldots \ldots+\int_7^8 e ^7 dt \right\}\)…
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