JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int {{x^5}{e^{ - 4{x^3}}}\,dx = \frac{1}{{48}}{e^{ - 4{x^3}}}f\left( x \right) + C} \), where \(C\) is a constant of integration, then \(f(x)\) is equal to
- A \(-2x^3 -1\)
- B \(-4x^3 -1\)
- C \(-2x^3 +1\)
- D \(4x^3+1\)
Answer & Solution
Correct Answer
(B) \(-4x^3 -1\)
Step-by-step Solution
Detailed explanation
\(\int x^{5} \cdot e^{-4 x^{2}} d x\) \(=\int x^{2} \cdot x^{3} e^{-4 x^{3}} d x\) \(-4 x^{3}=t\) \(-12 x^{2} d x=d t\) \(=\frac{-1}{12} \int-\frac{t}{4} e^{t} d t\) \(=\frac{1}{48} \int t e^{t} d t\) \(=\frac{1}{48} \mathrm{te}^{\mathrm{t}}-1 . \mathrm{e}^{t}+\mathrm{c}\)…
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