The value of the integral ∫13(x-2)^4sin^3(x-2)+(x-2)^2019+1dx is

Correct answer: (B) 2. ∫13(x-2)^4sin^3(x-2)+(x-2)^2019+1dx x-2=t⇒dx=dt ∫-11t^4sin^3t+t^2019+1dt=∫-11dt=t-11=2

Q43 / 91 · KVPY 2020 (SB/SX) · Mathematics

The value of the integral
$\int_{1}^{3} ((x - 2)^{4} \sin^{3} (x - 2) + (x - 2)^{2019} + 1) d x$ is

\int_{1}^{3} ((x - 2)^{4} \sin^{3} (x - 2) + (x - 2)^{2019} + 1) d x

x - 2 = t \Rightarrow d x = d t

{\int_{- 1}^{1} (t^{4} \sin^{3} t + t^{2019} + 1) d t = \int_{- 1}^{1} d t = t]}_{- 1}^{1} = 2

Hence, the correct answer is option: 'B'.