What is the Areas topic in JEE Mathematics?

The Areas topic is part of Applications of Integrals. It deals with calculating the area enclosed by curves, straight lines, coordinate axes, and other boundaries using definite integration.

Which concepts are important for JEE Areas questions?

Important concepts include the area under a curve, area between two curves, intersection points, definite integrals, symmetry, piecewise integration, and choosing whether to integrate with respect to x or y.

How many questions are asked from Areas in JEE Main?

JEE Main generally includes zero to one direct question from Areas or Applications of Integrals. The topic may contribute up to four marks when tested directly.

Are Areas questions important for JEE Advanced?

Yes. JEE Advanced may combine Areas with definite integration, coordinate geometry, functions, and inequalities. These questions often require students to identify boundaries and divide the region into multiple parts.

How do I find the area between two curves?

First, calculate the intersection points of the curves. When integrating with respect to x, subtract the lower curve from the upper curve. When integrating with respect to y, subtract the left boundary from the right boundary.

Why should I draw a graph before solving an Areas question?

A rough graph helps identify the required region, intersection points, integration limits, and relative positions of the curves. It also shows whether the integral must be divided into multiple intervals.

What are the common mistakes in JEE Areas questions?

Common mistakes include using incorrect limits, subtracting the curves in the wrong order, overlooking a change in sign, missing symmetry, and treating a negative definite integral as a negative geometric area.

How can I improve my speed in solving Areas questions?

Practise sketching standard curves, memorise important definite integration properties, calculate intersection points carefully, and use symmetry whenever possible. Solving previous-year questions also improves boundary identification and calculation speed.

JEE Areas Questions

Question 1 JEE Mains

Let $$A_{1}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,x+y=8$$ and y-axis that lies in the first quadrant. Let $$A_{2}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,y^{2}=x,x=2$$ and y-axis that lies in the first quadrant. Then $$A_{1}-A_{2}$$ is equal to

$$\frac{2}{3}(2\sqrt{2}+1)$$

$$\frac{2}{3}(\sqrt{2}+1)$$

$$\frac{2}{3}(4\sqrt{2}+1)$$

$$\frac{2}{3}(3\sqrt{2}+1)$$

Solution

Given curves for $$A_1$$:

$$y = x^2 + 2, \quad x + y = 8 \implies y = 8 - x, \quad x = 0 \quad (\text{y-axis})$$

Finding the point of intersection of $$y = x^2 + 2$$ and $$y = 8 - x$$:

$$x^2 + 2 = 8 - x \implies x^2 + x - 6 = 0 \implies (x+3)(x-2) = 0 \implies x = 2 \quad (\text{since } x \ge 0)$$

$$A_1 = \int_{0}^{2} \left[ (8 - x) - (x^2 + 2) \right] dx = \int_{0}^{2} (6 - x - x^2) dx$$

$$A_1 = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{2} = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}$$

Given curves for $$A_2$$:

$$y = x^2 + 2, \quad y^2 = x \implies y = \sqrt{x}, \quad x = 2, \quad x = 0$$

$$A_2 = \int_{0}^{2} \left[ (x^2 + 2) - \sqrt{x} \right] dx$$

$$A_2 = \left[ \frac{x^3}{3} + 2x - \frac{2}{3}x^{3/2} \right]_{0}^{2} = \frac{8}{3} + 4 - \frac{2}{3}(2\sqrt{2}) = \frac{20}{3} - \frac{4\sqrt{2}}{3}$$

$$A_1 - A_2 = \frac{22}{3} - \left( \frac{20}{3} - \frac{4\sqrt{2}}{3} \right) = \frac{2}{3} + \frac{4\sqrt{2}}{3} = \frac{2}{3}(2\sqrt{2} + 1)$$

Question 2 JEE Mains

Let the area of the region $$\{(x, y) : x - 2y + 4 \geq 0, \; x + 2y^2 \geq 0, \; x + 4y^2 \leq 8, \; y \geq 0\}$$ be $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime numbers. Then $$m + n$$ is equal to _______.

Solution

We need the area of the region defined by:

1) $$x - 2y + 4 \geq 0$$ (i.e., $$x \geq 2y - 4$$)

2) $$x + 2y^2 \geq 0$$ (i.e., $$x \geq -2y^2$$)

3) $$x + 4y^2 \leq 8$$ (i.e., $$x \leq 8 - 4y^2$$)

4) $$y \geq 0$$

Let us integrate with respect to $$y$$. From condition 3: $$x \leq 8 - 4y^2$$, so $$y \leq \sqrt{2}$$ (for $$x \geq 0$$, max).

The right boundary is $$x = 8 - 4y^2$$.

The left boundary is $$x = \max(2y - 4, -2y^2)$$.

For $$y \geq 0$$: When is $$2y - 4 \geq -2y^2$$?

$$2y^2 + 2y - 4 \geq 0$$, i.e., $$y^2 + y - 2 \geq 0$$, i.e., $$(y+2)(y-1) \geq 0$$. Since $$y \geq 0$$: $$y \geq 1$$.

So left boundary = $$-2y^2$$ for $$0 \leq y \leq 1$$ and $$2y - 4$$ for $$y \geq 1$$.

Upper limit from right boundary: $$8 - 4y^2 \geq 2y - 4$$ gives $$4y^2 + 2y - 12 \leq 0$$, $$2y^2 + y - 6 \leq 0$$, $$(2y - 3)(y + 2) \leq 0$$, so $$y \leq 3/2$$.

Also $$8 - 4y^2 \geq -2y^2$$ gives $$2y^2 \leq 8$$, $$y \leq 2$$.

Area = $$\int_0^1 (8 - 4y^2 - (-2y^2)) dy + \int_1^{3/2} (8 - 4y^2 - (2y - 4)) dy$$

$$= \int_0^1 (8 - 2y^2) dy + \int_1^{3/2} (12 - 4y^2 - 2y) dy$$

First integral: $$\left[8y - \frac{2y^3}{3}\right]_0^1 = 8 - \frac{2}{3} = \frac{22}{3}$$

Second integral: $$\left[12y - \frac{4y^3}{3} - y^2\right]_1^{3/2}$$

At $$y = 3/2$$: $$18 - \frac{4 \cdot 27/8}{3} - \frac{9}{4} = 18 - \frac{9}{2} - \frac{9}{4} = 18 - \frac{27}{4} = \frac{45}{4}$$

At $$y = 1$$: $$12 - \frac{4}{3} - 1 = 11 - \frac{4}{3} = \frac{29}{3}$$

Second integral = $$\frac{45}{4} - \frac{29}{3} = \frac{135 - 116}{12} = \frac{19}{12}$$

Total area = $$\frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}$$

$$m + n = 107 + 12 = 119$$.

The answer is $$\boxed{119}$$.

Question 3 JEE Mains

If the tangent at a point P on the parabola $$y^2 = 3x$$ is parallel to the line $$x + 2y = 1$$ and the tangents at the points Q and R on the ellipse $$\frac{x^2}{4} + \frac{y^2}{1} = 1$$ are perpendicular to the line $$x - y = 2$$, then the area of the triangle $$PQR$$ is:

$$\frac{9}{\sqrt{5}}$$

$$5\sqrt{3}$$

$$\frac{3}{2}\sqrt{5}$$

$$3\sqrt{5}$$

Solution

We need to find the area of triangle PQR. To determine point P on the parabola $$y^2 = 3x$$, note that the tangent at point $$(at^2, 2at)$$ on $$y^2 = 4ax$$ has slope $$\frac{1}{t}$$. Here $$4a = 3$$, so $$a = \frac{3}{4}$$. The line $$x + 2y = 1$$ has slope $$-\frac{1}{2}$$. Thus $$\frac{1}{t} = -\frac{1}{2}$$, giving $$t = -2$$. It follows that $$P = \left(\frac{3}{4} \times 4, 2 \times \frac{3}{4} \times (-2)\right) = (3, -3)$$.

Next, we find points Q and R on the ellipse $$\frac{x^2}{4} + y^2 = 1$$. Lines perpendicular to $$x - y = 2$$ (slope 1) have slope $$-1$$. The general tangent to the ellipse is given by $$y = mx \pm \sqrt{a^2m^2 + b^2}$$ where $$a^2 = 4, b^2 = 1$$. Hence the tangents with slope -1 are $$y = -x \pm \sqrt{4 + 1} = -x \pm \sqrt{5}$$.

For the tangent $$y = -x + \sqrt{5}$$, substituting into the ellipse equation yields

$$\frac{x^2}{4} + (-x + \sqrt{5})^2 = 1$$

$$\frac{x^2}{4} + x^2 - 2\sqrt{5}x + 5 = 1$$

$$\frac{5x^2}{4} - 2\sqrt{5}x + 4 = 0$$

$$5x^2 - 8\sqrt{5}x + 16 = 0$$

$$x = \frac{8\sqrt{5} \pm \sqrt{320 - 320}}{10} = \frac{8\sqrt{5}}{10} = \frac{4\sqrt{5}}{5}$$.

Then

$$y = -\frac{4\sqrt{5}}{5} + \sqrt{5} = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$$, so $$Q = \left(\frac{4\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right) = \left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$$. Similarly, for the tangent $$y = -x - \sqrt{5}$$ one finds $$R = \left(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$$.

Finally, the area of triangle PQR with vertices $$P(3,-3)$$, $$Q\left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$$, and $$R\left(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$$ is given by

Area $$= \frac{1}{2}|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)|$$

$$= \frac{1}{2}\left|3 \cdot \frac{2}{\sqrt{5}} + \frac{4}{\sqrt{5}}\left(-\frac{1}{\sqrt{5}} - (-3)\right) + \left(-\frac{4}{\sqrt{5}}\right)\left(-3 - \frac{1}{\sqrt{5}}\right)\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{4}{\sqrt{5}} \cdot \frac{-1 + 3\sqrt{5}}{\sqrt{5}} - \frac{4}{\sqrt{5}} \cdot \frac{-3\sqrt{5} - 1}{\sqrt{5}}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{4(-1+3\sqrt{5})}{5} + \frac{4(3\sqrt{5}+1)}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{-4+12\sqrt{5}+12\sqrt{5}+4}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{24\sqrt{5}}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{24}{\sqrt{5}}\right| = \frac{1}{2} \cdot \frac{30}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$$.

Therefore, the area is $$\boxed{3\sqrt{5}}$$, which corresponds to Option D.

Question 4 JEE Mains

What is the area bounded by $$y=x^3-7x^2+14x-8$$ and the x-axis in the interval $$[0,2]$$?

$$2$$

$$5/2$$

$$3$$

$$7/2$$

Solution

Factor the polynomial to find where the curve meets the x axis:

$$y = x^3 - 7x^2 + 14x - 8$$

$$y = (x - 1)(x - 2)(x - 4)$$

The required interval goes from zero to two.

The roots of the polynomial inside this range are at $$x = 1$$ and $$x = 2$$.

The curve changes sign at $$x = 1$$:

From zero to one the function is negative so we take the negative of the integral
From one to two the function is positive so we take the standard integral

Set up the area calculation:

$$\text{Area} = -\int_0^1 (x^3 - 7x^2 + 14x - 8) \, dx + \int_1^2 (x^3 - 7x^2 + 14x - 8) \, dx$$

Find the antiderivative function:

$$F(x) = \frac{x^4}{4} - \frac{7x^3}{3} + 7x^2 - 8x$$

Evaluate the first integral from zero to one:

$$-\left( \frac{1}{4} - \frac{7}{3} + 7 - 8 \right) = -\left( \frac{3 - 28 - 12}{12} \right) = \frac{37}{12}$$

Evaluate the second integral from one to two:

$$\left( 4 - \frac{56}{3} + 28 - 16 \right) - \left( \frac{1}{4} - \frac{7}{3} + 7 - 8 \right) = \left( 16 - \frac{56}{3} \right) - \left( -\frac{37}{12} \right)$$

$$= -\frac{8}{3} + \frac{37}{12} = \frac{-32 + 37}{12} = \frac{5}{12}$$

Sum both portions to get the total area:

$$\text{Total Area} = \frac{37}{12} + \frac{5}{12} = \frac{42}{12} = \frac{7}{2}$$

The correct option is D.

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