IIT-JEE 2009 · Paper 1 · Mathematics Q29

MathematicsMultiple correct+4 / -1Question 29 of 60

Area of the region bounded by the curve $y=e^x$ and lines $x=0$ and $y=e$ is

e-1

\displaystyle\int_1^e\ln(e+1-y)\,dy

e-\displaystyle\int_0^1e^x\,dx

\displaystyle\int_1^e\ln y\,dy

Answer and worked solution

Official answer: B, C, D

The area. The region lies between $y=e^x$ and $y=e$ for $0\le x\le1$ :
$\int_0^1(e-e^x)\,dx=e-(e-1)=1.$

(A) $e-1\ne1$ .
(B) With $u=e+1-y$ : $\int_1^e\ln u\,du=[u\ln u-u]_1^e=1$ . ✓
(C) $e-\int_0^1e^xdx=e-(e-1)=1$ . ✓
(D) In terms of $y$ : $x$ runs from 0 to $\ln y$ for $1\le y\le e$ , so the area is $\int_1^e\ln y\,dy=1$ . ✓

(B), (C), (D)

Topics:Area under a curve Properties of definite integrals

IIT-JEE 2009 · Paper 1, Mathematics question 29