Application of Integration
65 Questions
1996
JEE Advanced
Numerical
IIT-JEE 1996
Let ${A_n}$ be the area bounded by the curve $y = {\left( {\tan x} \right)^n}$ and the
lines $x=0,$ $y=0,$ and $x = {\pi \over 4}.$ Prove that for $n > 2,$
${A_n} + {A_{n - 2}} = {1 \over {n - 1}}$ and deduce ${1 \over {2n + 2}} < {A_n} < {1 \over {2n - 2}}.$
lines $x=0,$ $y=0,$ and $x = {\pi \over 4}.$ Prove that for $n > 2,$
${A_n} + {A_{n - 2}} = {1 \over {n - 1}}$ and deduce ${1 \over {2n + 2}} < {A_n} < {1 \over {2n - 2}}.$
Correct Answer: Solve it.
1995
JEE Advanced
Numerical
IIT-JEE 1995
Consider a square with vertices at $(1,1), (-1,1), (-1,-1)$ and $(1, -1)$. Let $S$ be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region $S$ and find its area.
Correct Answer: $${{16\sqrt 2 - 20} \over 3}$$
1994
JEE Advanced
Numerical
IIT-JEE 1994
In what ratio does the $x$-axis divide the area of the region
bounded by the parabolas $y = 4x - {x^2}$ and $y = {x^2} - x?$
bounded by the parabolas $y = 4x - {x^2}$ and $y = {x^2} - x?$
Correct Answer: $$121:4$$
1992
JEE Advanced
Numerical
IIT-JEE 1992
Sketch the region bounded by the curves $y = {x^2}$ and
$y = {2 \over {1 + {x^2}}}.$ Find the area.
$y = {2 \over {1 + {x^2}}}.$ Find the area.
Correct Answer: $$\left( {\pi - {2 \over 3}} \right)$$ sq. units
1991
JEE Advanced
Numerical
IIT-JEE 1991
If $'f$ is a continuous function with $\int\limits_0^x {f\left( t \right)dt \to \infty } $ as $\left| x \right| \to \infty ,$ then show that every line $y=mx$
intersects the curve ${y^2} + \int\limits_0^x {f\left( t \right)dt = 2!} $
intersects the curve ${y^2} + \int\limits_0^x {f\left( t \right)dt = 2!} $
Correct Answer: Solve it.
1991
JEE Advanced
Numerical
IIT-JEE 1991
Sketch the curves and identify the region bounded by
$x = {1 \over 2},x = 2,y = \ln \,x$ and $y = {2^x}.$ Find the area of this region.
$x = {1 \over 2},x = 2,y = \ln \,x$ and $y = {2^x}.$ Find the area of this region.
Correct Answer: $$\,{{4 - \sqrt 2 } \over {\log 2}} - {5 \over 2}\log 2 + {3 \over 2}$$
1990
JEE Advanced
Numerical
IIT-JEE 1990
Compute the area of the region bounded by the curves $\,y = ex\,\ln x$ and $y = {{\ln x} \over {ex}}$ where $ln$ $e=1.$
Correct Answer: $${{{e^2} - 5} \over {4e}}$$
1988
JEE Advanced
Numerical
IIT-JEE 1988
Find the area of the region bounded by the curve $C:y=$
$\tan x,$ tangent drawn to $C$ at $x = {\pi \over 4}$ and the $x$-axis.
$\tan x,$ tangent drawn to $C$ at $x = {\pi \over 4}$ and the $x$-axis.
Correct Answer: $${1 \over 2}\left[ {\log 2 - {1 \over 2}} \right]$$ sq. units
1987
JEE Advanced
Numerical
IIT-JEE 1987
Find the area bounded by the curves, ${x^2} + {y^2} = 25,\,4y = \left| {4 - {x^2}} \right|$ and $x=0$ above the $x$-axis.
Correct Answer: $$4 + 25{\sin ^{ - 1}}{4 \over 5}$$
1985
JEE Advanced
Numerical
IIT-JEE 1985
Sketch the region bounded by the curves $y = \sqrt {5 - {x^2}} $ and $y = \left| {x - 1} \right|$ and find its area.
Correct Answer: $${{5\pi - 2} \over 4}$$ sq. units
1984
JEE Advanced
Numerical
IIT-JEE 1984
Find the area of the region bounded by the $x$-axis and the curves defined by
$$y = \tan x, - {\pi \over 3} \le x \le {\pi \over 3};\,\,y = \cot x,{\pi \over 6} \le x \le {{3\pi } \over 2}$$
Correct Answer: $$\log {3 \over 2}$$ sq. units
1983
JEE Advanced
Numerical
IIT-JEE 1983
Find the area bounded by the $x$-axis, part of the curve $y = \left( {1 + {8 \over {{x^2}}}} \right)$ and
the ordinates at $x=2$ and $x=4$. If the ordinate at $x=a$ divides the area into two equal parts, find $a$.
the ordinates at $x=2$ and $x=4$. If the ordinate at $x=a$ divides the area into two equal parts, find $a$.
Correct Answer: $$a = 2\sqrt 2 $$
1982
JEE Advanced
MCQ
IIT-JEE 1982
The area bounded by the curves $y=f(x)$, the $x$-axis and the ordinates $x=1$ and $x=b$ is $(b-1)$ sin $(3b+4)$. Then $f(x)$ is
A.
$\left( {x - 1} \right)\cos \left( {3x + 4} \right)$
B.
$\sin \left( {3x + 4} \right)$
C.
$\sin \left( {3x + 4} \right) + 3\left( {x - 1} \right)\cos \left( {3x + 4} \right)$
D.
none of these
1982
JEE Advanced
Numerical
IIT-JEE 1982
For any real $t,\,x = {{{e^t} + {e^{ - t}}} \over 2},\,\,y = {{{e^t} - {e^{ - t}}} \over 2}$ is a point on the
hyperbola ${x^2} - {y^2} = 1$. Show that the area bounded by this hyperbola and the lines joining its centre to the points corresponding to ${t_1}$ and $-{t_1}$ is ${t_1}$.
hyperbola ${x^2} - {y^2} = 1$. Show that the area bounded by this hyperbola and the lines joining its centre to the points corresponding to ${t_1}$ and $-{t_1}$ is ${t_1}$.
Correct Answer: Solve it.
1981
JEE Advanced
Numerical
IIT-JEE 1981
Find the area bounded by the curve ${x^2} = 4y$ and the straight
Correct Answer: $${{9 \over 8}}$$ sq. units