The value of ∫ _ 0 ^ 1 4 x ^ 3 \ d ^ 2 d x ^ 2 (1 - x ^ 2 ) ^ 5 \ dx is ____________

Correct answer: 0 ∫ _ 0 ^ 1 4 x ^ 3 _ ⏟ d ^ 2 d x ^ 2 (1 - x ^ 2 ) ^ 5 _ ⏟ dx = [4 x ^ 3 (d)/(dx) (1 - x ^ 2 ) ^ 5 ] _ 0 ^ 1 - ∫ _ 0 ^ 1 12 x ^ 2 (d)/(dx) (1 - x ^ 2 ) ^ 5 dx = [4 x ^ 3 × 5 (1 - x ^ 2 ) ^ 4 (- 2x)] _ 0 ^ 1 - 12[ [ x ^ 2 (1 - x ^ 2 ) ^ 5 ] _ 0 ^ 1 - ∫ _ 0 ^ 1 2x (1 - x ^ 2 ) ^ 5 dx] = 0 - 0 - 12[0 - 0] + 12 ∫ _ 0 ^ 1 2x (1 - x ^ 2 ) ^ 5 dx = 12 × [- (1 - x ^ 2 ) ^ 6 6 ] _ 0 ^ 1 = 12[0 + (1)/(6)] = 2

JEE Advanced 2014Paper 1

Exam: JEE Advanced
Year: 2014
Paper: Paper 1
Subject: Mathematics
Chapter: Definite Integration

JEE Advanced 2014 · Paper 1 · Mathematics · Definite Integration

The value of ∫ _ 0 ^ 1 4 x ^ 3 \ d ^ 2 d x ^ 2 (1 - x ^ 2 ) ^ 5 \...

The value of

\mathop{∫}_{0}^{1}4{x}^{3}

\left\{\dfrac{{d}^{2}}{d{x}^{2}}{

\left(1 - {x}^{2}\right)}^{5}\right\}dx

is ____________

Correct answer

Step-by-step solution

\mathop{∫}_{0}^{1}\mathop{4{x}^{3}}_{⏟}\mathop{\dfrac{{d}^{2}}{d{x}^{2}}{\left(1 - {x}^{2}\right)}^{5}}_{⏟}dx

= {\left[4{x}^{3}\dfrac{d}{dx}{\left(1 - {x}^{2}\right)}^{5}\right]}_{0}^{1} - \mathop{∫}_{0}^{1}12{x}^{2}\dfrac{d}{dx}{\left(1 - {x}^{2}\right)}^{5}dx

= {\left[4{x}^{3} \times 5{\left(1 - {x}^{2}\right)}^{4}\left(- 2x\right)\right]}_{0}^{1} - 12\left[{\left[{x}^{2}{\left(1 - {x}^{2}\right)}^{5}\right]}_{0}^{1} - \mathop{∫}_{0}^{1}2x{\left(1 - {x}^{2}\right)}^{5}dx\right]

= 0 - 0 - 12\left[0 - 0\right] + 12\mathop{∫}_{0}^{1}2x{\left(1 - {x}^{2}\right)}^{5}dx

= 12 \times {\left[- \dfrac{{\left(1 - {x}^{2}\right)}^{6}}{6}\right]}_{0}^{1}

= 12\left[0 + \dfrac{1}{6}\right] = 2

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