JEE Mains Maths · Application of Integrals

Trigonometric, Exponential and Reciprocal Curves

Areas bounded by sine and cosine, exponential and log curves, or the curve y = k/x, where the answer carries a surd, e or a logarithm.

All Application of Integrals notes JEE Mains Maths strategy

Why this matters

Eighteen PYQs, fifteen of them multiple choice, and six from 2026. Eight use sin x and cos x, which cross where tan x = 1; four use eˣ, 2ˣ or a log curve; six bound the region by y = k/x or xy = k, so the answer has a log. Three ideas cover the page.

Concept 1 of 3: Sine and cosine

\sin x

and

\cos x

cross at

\frac\pi4+n\pi

, and a min or max of the two switches there. Their difference is

\sqrt2\sin\left(x-\frac\pi4\right)

, a single sine wave, so the area between consecutive crossings is fixed. Split at every crossing and at every zero of the curves.

Definition

$\sin x=\cos x$ at $x=\frac\pi4+n\pi$ .
$\sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right)$ ; $\sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right)$ .
One arch of $\sin x$ or $\cos x$ has area 2.

Between consecutive crossings

\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx=2\sqrt2

Worked example

Find the area between

y=\sin x

and

y=\cos x

for

0\le x\le\frac\pi2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q162Moderate

Example 1 · Application of Integrals · Trigonometric, Exponential and Reciprocal Curves

The area of the region

A =\left\{ (x,y):|\cos x - \sin x| \leq y \leq \sin x,0 \leq x \leq\frac{\pi}{2} \right\}

Area is not the signed integral

\int_0^{2\pi}\sin x\,dx=0

, but the area is 4. Split wherever the curve crosses the axis, or where the top and bottom swap, and add the pieces as positive numbers.

Drill 7 more on sine and cosine

Concept 2 of 3: Exponential and log curves

e^x

integrates to itself and

a^x

\frac{a^x}{\ln a}

\ln x

is the reflection of

e^x

y=x

, so a region beside a log curve is often easier with horizontal strips, where the boundary is

x=e^y

Definition

$\int e^{kx}\,dx=\frac{e^{kx}}k$ ; $\int a^x\,dx=\frac{a^x}{\ln a}$ .
$\int\ln x\,dx=x\ln x-x$ .
$y=\ln x$ is $x=e^y$ : the area between $\ln x$ , the x-axis and $x=e$ is $\int_0^1(e-e^y)\,dy=1$ .

Log and exponential

\int\ln x\,dx=x\ln x-x+C,\qquad\int a^x\,dx=\frac{a^x}{\ln a}+C

Worked example

Find the area between

y=e^x

and

y=x+1

for

0\le x\le1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q70Moderate

Example 2 · Application of Integrals · Trigonometric, Exponential and Reciprocal Curves

Let e be the base of natural logarithm and let

f:\{ 1,2,3,4\} \rightarrow\left\{ 1,e,e^{2},e^{3} \right\}

and

g:\left\{ 1,e,e^{2},e^{3} \right\}\rightarrow\left\{ 1,\frac{1}{2},\frac{1}{3},\frac{1}{4} \right\}

be two bijective functions such that

f

is strictly decreasing and g is strictly increasing. If

\phi(x) =\left\lbrack f^{- 1}\left\{ {\text{ }g}^{- 1}\left( \frac{1}{2} \right) \right\} \right\rbrack^{x}

, then the area of the region

R =\left\{ (x,y):x^{2}\leq y \leq\phi(x) \right.

0 \leq x \leq 1\}

is :

The ln a factor

\int_0^1 2^x\,dx=\frac1{\ln2}

, not 1. Only base e integrates to itself; any other base divides by its log.

Drill 3 more on exponential and log curves

Concept 3 of 3: Reciprocal curves and xy = k

xy=k

is the curve

y=\frac kx

, and its integral is

k\ln x

, so the answer has a log. Find where the hyperbola meets the other curve; the top usually switches to the hyperbola there, so split the interval at that point.

Definition

For $x>0$ , $xy=k$ is $y=\frac kx$ , and $\int_a^b\frac kx\,dx=k\ln\frac ba$ .
$(x+c)y=k$ is $y=\frac k{x+c}$ , with integral $k\ln(x+c)$ .
$xy\le k$ with no condition on x lets x run negative, where the region can be unbounded.

Under y = k/x

\int_a^b\frac{k}{x}\,dx=k\ln\frac{b}{a}

Worked example

Find the area between

y=\frac1x

and the line

x+y=\frac52

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q63Moderate

Example 3 · Application of Integrals · Trigonometric, Exponential and Reciprocal Curves

The area of the region

R=\left\{ (x,y):xy\leq 8,1 \leq y\leq x^{2},x\geq 0 \right\}

State the quadrant

Without

x\ge0

, a region such as

xy\le k

1\le y\le x^2

includes every

x\le-1

, where

xy\le0\le k

always holds, so it is unbounded. The printed answers assume

x\ge0

Drill 5 more on reciprocal curves and xy = k

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Sine and cosine
Between consecutive crossings
$\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx=2\sqrt2$
Exponential and log curves
Log and exponential
$\int\ln x\,dx=x\ln x-x+C,\qquad\int a^x\,dx=\frac{a^x}{\ln a}+C$
Reciprocal curves and xy = k
Under y = k/x
$\int_a^b\frac{k}{x}\,dx=k\ln\frac{b}{a}$

Watch out for (3)

Area is not the signed integral
→ Sine and cosine
The ln a factor
→ Exponential and log curves
State the quadrant
→ Reciprocal curves and xy = k

Test yourself on Application of Integrals

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