The value of the integral ∫ x^ (1)/(3) (1 - x ) ^ 3 dx is equal to (where c & is the constant of integration)

Correct answer: B. 6((x(4)/(3))/(8) - (3)/(11)x(11)/(6) + (3)/(14)x(7)/(3) - (1)/(17)x(17)/(6)) + c Let, I = ∫ x^ (1)/(3) (1 - x ) ^ 3 dx Let, x = t^ 6 ⇒ dx = 6t^ 5 dt I = ∫ t^ 2 (1 - t^ 3 ) ^ 3 6t^ 5 dt I = 6∫ t^ 7 (1 - 3t^ 3 + 3t^ 6 - t^ 9 )dt = 6∫ (t^ 7 - 3t^ 10 + 3t^ 13 - t^ 16 )dt I = 6((t8)/(8) - 3t^ 11 11 + 3t^ 14 14 - (t17)/(17)) + c I = 6((x(4)/(3))/(8) - (3)/(11)x(11)/(6) + (3)/(14)x(7)/(3) - (1)/(17)x(17)/(6)) + c

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of the integral ∫ x^ (1)/(3) (1 - x ) ^ 3 dx is equal to...

The value of the integral

∫ x^{\dfrac{1}{3}}{\left(1 - \sqrt{x}\right)}^{3}dx

is equal to (where

c

& is the constant of integration)

Options

A $6\left(\dfrac{x^{\dfrac{4}{3}}}{8} + \dfrac{3}{11}x^{\dfrac{11}{6}} + \dfrac{3}{14}x^{\dfrac{7}{3}} + \dfrac{1}{17}x^{\dfrac{17}{6}}\right) + c$
B $6\left(\dfrac{x\dfrac{4}{3}}{8} - \dfrac{3}{11}x\dfrac{11}{6} + \dfrac{3}{14}x\dfrac{7}{3} - \dfrac{1}{17}x\dfrac{17}{6}\right) + c$
C $2\left(\dfrac{x^{\dfrac{4}{3}}}{8} - \dfrac{3}{11}x^{\dfrac{4}{6}} - \dfrac{3}{14}x^{\dfrac{7}{3}} - \dfrac{1}{17}x^{\dfrac{17}{6}}\right) + c$
D $2\left(\dfrac{x^{4}}{8} - \dfrac{3}{11}x^{11} - \dfrac{3}{14}x^{7} - \dfrac{1}{17}x^{17}\right) + c$

Correct answer

B. $6\left(\dfrac{x\dfrac{4}{3}}{8} - \dfrac{3}{11}x\dfrac{11}{6} + \dfrac{3}{14}x\dfrac{7}{3} - \dfrac{1}{17}x\dfrac{17}{6}\right) + c$

Step-by-step solution

Let,

I = ∫ x^{\dfrac{1}{3}}{\left(1 - \sqrt{x}\right)}^{3}dx

Let,

x = t^{6}⇒ dx = 6t^{5}dt

I = ∫ t^{2}{\left(1 - t^{3}\right)}^{3}6t^{5}dt

I = 6∫ t^{7}\left(1 - 3t^{3} + 3t^{6} - t^{9}\right)dt

= 6∫ \left(t^{7} - 3t^{10} + 3t^{13} - t^{16}\right)dt

I = 6\left(\dfrac{t8}{8} - \dfrac{3t^{11}}{11} + \dfrac{3t^{14}}{14} - \dfrac{t17}{17}\right) + c

I = 6\left(\dfrac{x\dfrac{4}{3}}{8} - \dfrac{3}{11}x\dfrac{11}{6} + \dfrac{3}{14}x\dfrac{7}{3} - \dfrac{1}{17}x\dfrac{17}{6}\right) + c

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