Question

\(\rm\int\limits_0^1 4\sqrt{1-x^2}dx\) = 
1. 1
2. π 
3. π/2
4. 4π 

Answer 1

Correct Answer - Option 2 : π 

Concept: 

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

 

Calculation:

I = \(\rm\int4\sqrt{1-x^2}dx\)

Let x = cos t 

⇒ dx = -sin t dt

I = \(\rm\int4\sqrt{1-\cos^2t}(-\sin t )dt \)

I = \(\rm-4\int\sin^2tdt\)

I = \(\rm-2\int1-\cos 2tdt\)

I = \(\rm-2\left[t-{\sin 2t\over2}\right]\)

I = sin 2t - 2t

I = sin 2(cos-1 x) - 2cos-1 x

Putting the limits, we get

I = \(\rm \left[\sin 2(\cos^{-1} x) - 2\cos^{-1} x\right]_0^1\)

I = \(\rm \left[\sin 2(\cos^{-1} 1) - 2\cos^{-1} 1\right]- \left[\sin 2(\cos^{-1} 0) - 2\cos^{-1} 0\right]\)

I = sin 2(0) - 2(0) - (sin 2(π/2) - 2(π/2))

I = π