NCERT Class 12 Mathematics — Application of Integrals, Exercise 8.1. All 4 questions solved.

This is the shortest exercise in the Class 12 syllabus — four questions — and every one of them rests on the same two formulas:

$$\text{Area} = \int_a^b y\,dx = \int_a^b f(x)\,dx \qquad \text{(vertical strips)}$$

$$\text{Area} = \int_c^d x\,dy = \int_c^d \phi(y)\,dy \qquad \text{(horizontal strips)}$$

Key insight. Choosing between the two formulas is the whole decision. Use vertical strips when the region is bounded above and below by curves and left and right by ordinates; use horizontal strips when it is bounded left and right by curves and above and below by lines $y = c$ and $y = d$. Question 4 is bounded by the $y$-axis and the line $y = 3$, so horizontal strips are the natural choice and vertical strips make it awkward.

The second habit worth forming now: sketch first. An integral cannot tell you that part of the region lies below the axis; only the sketch can.

Question 1

Find the area of the region bounded by the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.

Solution. The ellipse has $a^2 = 16$ and $b^2 = 9$, so $a = 4$ and $b = 3$. It is symmetric about both axes, so the total area is four times the area in the first quadrant. Solving for $y$ in that quadrant:

$$y = \frac{3}{4}\sqrt{16 – x^2}$$

$$\text{Area} = 4\int_0^4 \frac34\sqrt{16-x^2}\,dx = 3\left[\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\frac{x}{4}\right]_0^4$$

At $x = 4$ the first term vanishes and $\sin^{-1}1 = \tfrac{\pi}{2}$, so the bracket is $4\pi$.

Using symmetry rather than integrating from $-4$ to $4$ avoids the square root’s sign ambiguity on the negative half.

Question 2

Find the area of the region bounded by the ellipse $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$.

Solution. Here $a = 2$ and $b = 3$. The same computation with $y = \tfrac{3}{2}\sqrt{4-x^2}$ gives

$$4\int_0^2 \frac32\sqrt{4-x^2}\,dx = 6\left[\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2}\right]_0^2 = 6\pi$$

Both questions confirm the general result $\pi a b$ derived in the chapter: the area of an ellipse is $\pi$ times the product of its semi-axes, which reduces to $\pi r^2$ when $a = b = r$.

Choose the correct answer in the following Exercises 3 and 4.

Question 3

Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is

Solution. The circle has radius $2$. The lines $x = 0$ and $x = 2$ are the $y$-axis and the vertical tangent at the far right, so together with the first quadrant they bound exactly one quarter of the disc:

$$\int_0^2\sqrt{4-x^2}\,dx = \left[\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2}\right]_0^2 = 2 \cdot \frac{\pi}{2} = \pi$$

which agrees with $\tfrac14\pi r^2 = \tfrac14\pi(4)$. Recognising the quarter-disc lets you check the integral rather than trust it.

Question 4

Area of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is

Solution. The boundaries are the $y$-axis on the left and the horizontal line $y = 3$ on top, so horizontal strips are the right choice. Write $x$ in terms of $y$:

$$x = \frac{y^2}{4}$$

$$\text{Area} = \int_0^3 x\,dy = \int_0^3\frac{y^2}{4}\,dy = \frac{1}{4} \cdot \frac{27}{3}$$

Option (D) is what you get by using $\int_0^3 y\,dy$ — integrating the wrong variable’s expression.

Common mistakes

• Integrating from $-a$ to $a$ across a square root. $\sqrt{a^2-x^2}$ gives only the upper half of the ellipse or circle, so integrating it across the full width returns half the area, not all of it. Use symmetry and multiply.
• Choosing vertical strips when the region is described by $y$-boundaries. Question 4 is bounded by the $y$-axis and $y = 3$; forcing $\int y\,dx$ on it means splitting the region and subtracting a rectangle.
• Forgetting that area is unsigned. None of the four regions here dips below the axis, but the habit of checking is worth forming — the Miscellaneous Exercise depends on it.
• Misreading the semi-axes. In $\tfrac{x^2}{16} + \tfrac{y^2}{9} = 1$ the semi-axes are $4$ and $3$, not $16$ and $9$. The denominators are $a^2$ and $b^2$.

Practise next

• Miscellaneous Exercise on Chapter 8 — areas under $y = x^n$, a modulus graph, and regions that cross the axis.
• Exercise 7.10 — the properties of definite integrals used to handle the symmetry in these regions.