The integral I = ∫ 2 ^ ( 2 ^ x + x) dx = λ. ( 2 ^ 2 ^ x ) + C (where, C is the constant of integration). Then the value of& λ and is equal to

Correct answer: C. (1)/( 2) Given, I = ∫ 2 ^ 2 ^ x . 2 ^ x dx Let, 2 ^ 2 ^ x = t ⇒ 2 ^ 2 ^ x . 2 ^ x . ( 2) ^ 2 dx = dt ⇒ 2 ^ 2 ^ x . 2 ^ x dx = dt ( 2) ^ 2 Thus, I = ∫ dt ( 2) ^ 2 = t ( 2) ^ 2 + C = 2 ^ 2 ^ x ( 2) ^ 2 + C ∴ λ = 1 ( 2) ^ 2 ⇒ λ = (l)/( 2)

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The integral I = ∫ 2 ^ ( 2 ^ x + x) dx = λ. ( 2 ^ 2 ^ x ) + C...

The integral

I = ∫ {2}^{\left({2}^{x} + x\right)}dx = \lambda. \left({2}^{{2}^{x}}\right) + C

(where,

C

is the constant of integration). Then the value of&

\sqrt{\lambda} \text{ and } is equal to

Options

A $\dfrac{1}{\ln 4}$
B\dfrac{1}{{ $\left(\ln 2\right)}^{2}}$
C $\dfrac{1}{\ln 2}$
D\dfrac{1}{{ $\left(\ln 4\right)}^{2}}$

Correct answer

C. $\dfrac{1}{\ln 2}$

Step-by-step solution

Given,

I = ∫ {2}^{{2}^{x}}{. 2}^{x}dx

Let,

{2}^{{2}^{x}} = t ⇒ {2}^{{2}^{x}}{. 2}^{x}. {\left(\ln 2\right)}^{2}dx = dt

⇒ {2}^{{2}^{x}}{. 2}^{x}dx = \dfrac{dt}{{\left(\ln 2\right)}^{2}}

Thus,

I = ∫ \dfrac{dt}{{\left(\ln 2\right)}^{2}} = \dfrac{t}{{\left(\ln 2\right)}^{2}} + C

= \dfrac{{2}^{{2}^{x}}}{{\left(\ln 2\right)}^{2}} + C

∴ \lambda = \dfrac{1}{{\left(\ln 2\right)}^{2}} ⇒ \sqrt{\lambda} = \dfrac{l}{\ln 2}

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