Applications of Integrals JEE Notes PDF: Download Now

Applications of Integrals JEE Notes give you exactly what JEE asks: finding the area under curves, between curves and a handful of related results. Use these crisp points for the final 48-hour run-through.

Applications of Integrals JEE Notes: Important Concepts

The chapter converts geometric area problems into definite integrals. You integrate the upper curve minus the lower curve, or the right curve minus the left, over the correct limits. JEE almost always keeps the region in 2-D Cartesian coordinates, yet questions may mix straight lines with conics, or ask for the least positive area when curves intersect more than twice.

• Bounded region: A finite closed area; integration limits must be the actual points of intersection.
• Sign sense: Area is always taken as positive. If an integral comes out negative, take its absolute value.
• Choice of axis: If the region is better split vertically, use $$\displaystyle \int_{x_1}^{x_2}(y_{\text{top}}-y_{\text{bottom}})\,dx$$. For horizontal strips, use $$\displaystyle \int_{y_1}^{y_2}(x_{\text{right}}-x_{\text{left}})\,dy$$.
• Piece-wise integration: When the bounding curves cross inside the interval, break the integral at each intersection.
• Average value: For a continuous function $$f(x)$$ on $$[a,b]$$, the average or mean value is $$\displaystyle \frac{1}{b-a}\int_a^b f(x)\,dx$$.

Area Under a Curve and Between Two Curves

Standard Procedure

1. Sketch or mentally picture the curves: avoids wrong limits.
2. Find intersection points by solving equations of the curves.
3. Decide strip orientation: vertical for functions $$y=f(x)$$, horizontal if $$x=g(y)$$ is easier.
4. Set up the definite integral: upper minus lower (vertical) or right minus left (horizontal).
5. Integrate and take the absolute value.

Worked Example 1 (Vertical Strips)

Find the area enclosed by $$y=x^2$$, $$y=2x$$.

Intersection: $$x^2=2x \Rightarrow x=0,\,2$$.

Upper curve in $$[0,2]$$ is $$y=2x$$; lower is $$x^2$$.

$$ \begin{aligned} A &= \int_0^2 (2x-x^2)\,dx \\ &= \left[x^2-\frac{x^3}{3}\right]_0^2 \\ &= \left(4-\frac{8}{3}\right)-0 \\ &= \frac{4}{3}. \end{aligned} $$

Area = $$\displaystyle \frac{4}{3}$$ square units.

Worked Example 2 (Horizontal Strips)

Region bounded by $$x=y^2$$, $$x=4$$, $$y=0$$.

Horizontal strip: right curve $$x=4$$, left $$x=y^2$$. Limits on $$y$$: from $$0$$ to $$2$$ (because $$y^2=4$$).

$$ \begin{aligned} A &= \int_0^2 (4-y^2)\,dy \\ &= \left[4y-\frac{y^3}{3}\right]_0^2 \\ &= 8-\frac{8}{3} \\ &= \frac{16}{3}. \end{aligned} $$

Area = $$\displaystyle \frac{16}{3}$$ square units.

After mastering the setup, test yourself on mixed-curve questions inside the JEE Questions so that the process becomes reflexive.

Areas Involving Standard Curves: Lines, Parabolas, Circles and Ellipses

Parabola with a Line

When a straight line cuts a parabola, JEE likes symmetric regions. Example: area enclosed between $$y^2=4ax$$ and $$y=mx$$ turns out proportional to $$a^2/m^3$$ after elimination.

Circle Segments and Sectors

• Sector of angle $$\theta$$ in a circle of radius $$r$$: $$\displaystyle \frac{1}{2}r^2\theta$$ (with $$\theta$$ in radians).
• Area of a segment = sector area minus triangle area.
• If the question gives polar form $$r=f(\theta)$$, use $$\displaystyle A=\frac{1}{2}\int_{\theta_1}^{\theta_2}r^2\,d\theta$$.

Ellipse

Complete ellipse area: $$\pi ab$$. For a portion bounded by a line or another curve, convert to parametric $$x=a\cos t$$, $$y=b\sin t$$ and integrate.

Worked Example 3 (Parabola & Line)

Find the area common to $$y=x$$ and $$y^2=4x$$.

Intersection: Substitute $$y=x$$ into the parabola: $$x^2=4x \Rightarrow x=0,4$$. The region lies in the first quadrant.

Since $$y=x$$ and $$y=2\sqrt{x}$$ are the upper and lower boundaries respectively only after checking their relative positions, the area is

$$ \begin{aligned} A &= \int_0^4 (2\sqrt{x}-x)\,dx \\ &= \left[\frac{4}{3}x^{3/2}-\frac{x^2}{2}\right]_0^4 \\ &= \frac{32}{3}-8 \\ &= \frac{8}{3}. \end{aligned} $$

Area = $$\displaystyle \frac{8}{3}$$ square units.

Keep a personal list of such standard results in your pocket notebook or use the downloadable JEE Formula Sheets for one-minute recall.

Integral as Accumulator: Average Value, Mean Value Theorem & Centroid Overview

Average (Mean) Value

$$ \text{Mean value of } f(x) \text{ on } [a,b] = \frac{1}{b-a}\int_a^b f(x)\,dx $$

Mean Value Theorem for Integrals

If $$f$$ is continuous on $$[a,b]$$, there exists $$c\in(a,b)$$ such that $$\displaystyle \int_a^b f(x)\,dx=f(c)(b-a)$$. JEE may hide this inside a statement-based question.

Centroid of a Plane Lamina (Quick Glance)

Though rare, the formulas are useful:

Worked Example 4 (Average Value)

For $$f(x)=\cos x$$ on $$[0,\pi/2]$$, find the average value.

$$ \begin{aligned} \text{Avg} &= \frac{1}{\pi/2-0}\int_0^{\pi/2}\cos x\,dx \\ &= \frac{2}{\pi}\left[\sin x\right]_0^{\pi/2} \\ &= \frac{2}{\pi}(1-0) \\ &= \frac{2}{\pi}. \end{aligned} $$

Average value = $$\displaystyle \frac{2}{\pi}$$.

Important Formulas and Results at a Glance

$$ A_{\text{between curves}} = \int_a^b [f(x)-g(x)]\,dx, \qquad f(x)\ge g(x)\quad\forall x\in[a,b] $$

If curves cross inside $$[a,b]$$, split at each intersection and add absolute areas.

JEE Important Points, Common Mistakes and Quick Revision

• Sketch first. A 15-second drawing prevents wrong limits, the #1 error in past JEE attempts.
• Check crossings. If the relative position flips, split the integral. Missing this costs full marks.
• Units slip. Keep answers in square units; JEE deducts for missing dimension in integer-type paper.
• Sign trap in horizontal strips. Always right minus left; not upper minus lower.
• Polar to Cartesian mix-up. Use the $$\displaystyle \frac{1}{2}r^2\,d\theta$$ formula only when the curve is in polar form.
• Final minute hack: memorize that the area between $$y=x$$ and $$y=x^2$$ from $$0$$ to $$1$$ is $$\displaystyle \frac{1}{6}$$: it appears in many variations.

Revisit solved numerical patterns from the last decade using JEE Mains Previous Papers; spotting the same curve pair twice is free score.