NCERT Exemplar Solutions Class 12 Maths Application of Integrals — Short Answer Q1–15

NCERT Exemplar Problems, Class 12 Mathematics — Application of Integrals, Short Answer questions 1 to 15. All 15 questions solved.

Fifteen area questions, and every one of them runs on the same three moves: find where the curves meet, decide which curve is on top (or to the right), and integrate the difference between those limits. What changes from question to question is only which of the three moves is the hard one — in Question 3 it is solving a cubic for the meeting point, in Question 15 it is noticing that integrating along $y$ saves splitting the region in two.

A note on the syllabus. The current NCERT textbook chapter teaches only the area under a single curve; the section on the area between two curves was removed in the rationalised edition. Questions 1–7 and 13–15 are of that kind. They remain squarely in the JEE Main and JEE Advanced syllabuses, which both name “areas of regions bounded by simple curves”, so they are worth working whichever examination you are preparing for.

Key insight. Area is always $\int(\text{upper} – \text{lower})\,dx$ or $\int(\text{right} – \text{left})\,dy$ between the meeting points. Before integrating, ask which variable makes the region a single strip. When the boundary curves are written as $x$ in terms of $y$ — a sideways parabola, a line given as $x = \ldots$ — integrating along $y$ usually does.

Question 1

Find the area of the region bounded by the curves $y^2 = 9x$ and $y = 3x$.

Solution. The parabola opens to the right; the line passes through the origin. They meet where $(3x)^2 = 9x$, that is $9x^2 = 9x$, so $x = 0$ or $x = 1$ — the points $(0, 0)$ and $(1, 3)$.

Between these, the region lies above the $x$-axis, with the parabola’s upper branch $y = 3\sqrt{x}$ on top (at $x = \frac14$, $3\sqrt x = \frac32$ while $3x = \frac34$). So

$$A = \int_0^1 \left(3\sqrt{x} – 3x\right)dx = \left[2x^{3/2} – \frac{3x^2}{2}\right]_0^1 = 2 – \frac32$$

Question 2

Find the area of the region bounded by the parabolas $y^2 = 2px$ and $x^2 = 2py$.

Solution. Take $p > 0$. The parabolas meet where $x = \dfrac{y^2}{2p}$ and $x^2 = 2py$: substituting, $\dfrac{y^4}{4p^2} = 2py$, so $y = 0$ or $y^3 = 8p^3$, giving $y = 2p$. The meeting points are $(0, 0)$ and $(2p, 2p)$.

On $0 \le x \le 2p$ the rightward parabola $y = \sqrt{2px}$ lies above the upward one $y = \dfrac{x^2}{2p}$, so

$$A = \int_0^{2p}\left(\sqrt{2p}\,x^{1/2} – \frac{x^2}{2p}\right)dx = \left[\frac{2\sqrt{2p}}{3}x^{3/2} – \frac{x^3}{6p}\right]_0^{2p} = \frac{2\sqrt{2p}\,(2p)^{3/2}}{3} – \frac{8p^3}{6p}$$

The first term is $\frac{2}{3}(2p)^2 = \frac{8p^2}{3}$ and the second is $\frac{4p^2}{3}$.

Worth remembering. For $y^2 = 4ax$ and $x^2 = 4by$ the area is $\frac{16ab}{3}$. Here $4a = 4b = 2p$, so $a = b = \frac p2$ and $\frac{16}{3}\cdot\frac{p^2}{4} = \frac{4p^2}{3}$ — a quick check on the integration.

Question 3

Find the area of the region bounded by the curve $y = x^3$ and $y = x + 6$ and $x = 0$.

Solution. The curve and the line meet where $x^3 = x + 6$, i.e. $x^3 – x – 6 = 0$. Trying small integers, $x = 2$ works, and dividing out gives $(x – 2)(x^2 + 2x + 3) = 0$. The quadratic has discriminant $4 – 12 < 0$, so $x = 2$ is the only meeting point, at $(2, 8)$.

The region is closed on the left by $x = 0$. On $[0, 2]$ the line is above the cubic (at $x = 0$ it is at $6$, the cubic at $0$), so

$$A = \int_0^2\left(x + 6 – x^3\right)dx = \left[\frac{x^2}{2} + 6x – \frac{x^4}{4}\right]_0^2 = 2 + 12 – 4$$

Question 4

Find the area of the region bounded by the curve $y^2 = 4x$ and $x^2 = 4y$.

Solution. Substituting $y = \frac{x^2}{4}$ into $y^2 = 4x$ gives $\frac{x^4}{16} = 4x$, so $x = 0$ or $x^3 = 64$, $x = 4$. The parabolas meet at $(0, 0)$ and $(4, 4)$, and between them $y = 2\sqrt x$ lies above $y = \frac{x^2}{4}$:

$$A = \int_0^4\left(2\sqrt{x} – \frac{x^2}{4}\right)dx = \left[\frac43 x^{3/2} – \frac{x^3}{12}\right]_0^4 = \frac{32}{3} – \frac{64}{12} = \frac{32}{3} – \frac{16}{3}$$

This is Question 2 with $2p = 4$, and $\frac{4p^2}{3} = \frac{4\cdot 4}{3}$ agrees.

Question 5

Find the area of the region included between $y^2 = 9x$ and $y = x$.

Solution. They meet where $x^2 = 9x$: at $x = 0$ and $x = 9$, the points $(0,0)$ and $(9, 9)$. Between them the parabola’s upper branch $y = 3\sqrt x$ is above the line (at $x = 1$, $3$ against $1$):

$$A = \int_0^9\left(3\sqrt{x} – x\right)dx = \left[2x^{3/2} – \frac{x^2}{2}\right]_0^9 = 2(27) – \frac{81}{2} = 54 – \frac{81}{2}$$

Question 6

Find the area of the region enclosed by the parabola $x^2 = y$ and the line $y = x + 2$.

Solution. The meeting points satisfy $x^2 = x + 2$, so $x^2 – x – 2 = (x – 2)(x + 1) = 0$, giving $x = -1$ and $x = 2$. Between them the line is above the parabola (at $x = 0$: $2$ against $0$):

$$A = \int_{-1}^{2}\left(x + 2 – x^2\right)dx = \left[\frac{x^2}{2} + 2x – \frac{x^3}{3}\right]_{-1}^{2} = \left(2 + 4 – \frac83\right) – \left(\frac12 – 2 + \frac13\right) = \frac{10}{3} + \frac76$$

A shortcut to check with. The area between $y = x^2$ and a line meeting it at $x = \alpha$ and $x = \beta$ is always $\frac{(\beta – \alpha)^3}{6}$. Here $\frac{3^3}{6} = \frac92$.

Question 7

Find the area of region bounded by the line $x = 2$ and the parabola $y^2 = 8x$.

Solution. The line $x = 2$ cuts the parabola at $y^2 = 16$, i.e. at $(2, \pm 4)$. The region is symmetric about the $x$-axis, so take twice the part above it:

$$A = 2\int_0^2 \sqrt{8x}\,dx = 2\cdot 2\sqrt2\left[\frac23 x^{3/2}\right]_0^2 = 4\sqrt2 \cdot \frac23 \cdot 2\sqrt2 = \frac{32}{3}$$

Why the factor 2 is needed. $y = \sqrt{8x}$ is only the upper half of the parabola. Integrating it alone gives $\frac{16}{3}$, the area above the axis — the most common way to lose half the marks on this question.

Question 8

Sketch the region $\left\{(x, 0) : y = \sqrt{4 – x^2}\right\}$ and $x$-axis. Find the area of the region using integration.

Solution. The set as printed is garbled — it collects points of the form $(x, 0)$, which lie on the axis and have no area. The intended region is the one bounded by $y = \sqrt{4 – x^2}$ and the $x$-axis, and that is what the key answers.

Squaring, $x^2 + y^2 = 4$ with $y \ge 0$: the upper half of the circle of radius $2$ centred at the origin. It meets the $x$-axis at $x = \pm 2$. Using the standard result

$$\int\sqrt{a^2 – x^2}\,dx = \frac{x}{2}\sqrt{a^2 – x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C,$$

$$A = \int_{-2}^{2}\sqrt{4 – x^2}\,dx = \left[\frac x2\sqrt{4 – x^2} + 2\sin^{-1}\frac x2\right]_{-2}^{2} = 2\cdot\frac\pi2 – 2\cdot\left(-\frac\pi2\right)$$

As it must be: half of $\pi r^2 = 4\pi$.

Question 9

Calculate the area under the curve $y = 2\sqrt{x}$ included between the lines $x = 0$ and $x = 1$.

Solution. The curve is non-negative on $[0, 1]$, so the area is the integral itself:

$$A = \int_0^1 2\sqrt{x}\,dx = \left[\frac43 x^{3/2}\right]_0^1$$

Question 10

Using integration, find the area of the region bounded by the line $2y = 5x + 7$, $x$-axis and the lines $x = 2$ and $x = 8$.

Solution. The line is $y = \frac{5x + 7}{2}$, which crosses the $x$-axis at $x = -\frac75$ — to the left of $x = 2$. So on $[2, 8]$ it is entirely above the axis, and no part of the region needs its sign flipped:

$$A = \int_2^8\frac{5x + 7}{2}\,dx = \frac12\left[\frac{5x^2}{2} + 7x\right]_2^8 = \frac12\left[(160 + 56) – (10 + 14)\right] = \frac12\cdot 192$$

Check by geometry. The region is a trapezium with parallel sides $y(2) = \frac{17}{2}$ and $y(8) = \frac{47}{2}$, and width $6$: $\frac12\left(\frac{17}{2} + \frac{47}{2}\right)\cdot 6 = 96$.

Question 11

Draw a rough sketch of the curve $y = \sqrt{x – 1}$ in the interval $[1, 5]$. Find the area under the curve and between the lines $x = 1$ and $x = 5$.

Solution. The curve is the upper half of the parabola $y^2 = x – 1$, whose vertex is at $(1, 0)$; it rises to $(5, 2)$. Substituting $t = x – 1$:

$$A = \int_1^5\sqrt{x – 1}\,dx = \left[\frac23 (x – 1)^{3/2}\right]_1^5 = \frac23\cdot 4^{3/2} = \frac23\cdot 8$$

Question 12

Determine the area under the curve $y = \sqrt{a^2 – x^2}$ included between the lines $x = 0$ and $x = a$.

Solution. This is the quarter of the circle $x^2 + y^2 = a^2$ in the first quadrant. With the same standard integral as Question 8:

$$A = \int_0^a\sqrt{a^2 – x^2}\,dx = \left[\frac x2\sqrt{a^2 – x^2} + \frac{a^2}{2}\sin^{-1}\frac xa\right]_0^a = \frac{a^2}{2}\cdot\frac\pi2$$

Question 13

Find the area of the region bounded by $y = \sqrt{x}$ and $y = x$.

Solution. They meet where $\sqrt x = x$, i.e. $x = 0$ and $x = 1$. On $(0, 1)$ the root is the larger ($\sqrt{0.25} = 0.5 > 0.25$):

$$A = \int_0^1\left(\sqrt{x} – x\right)dx = \left[\frac23 x^{3/2} – \frac{x^2}{2}\right]_0^1 = \frac23 – \frac12$$

Question 14

Find the area enclosed by the curve $y = -x^2$ and the straight line $x + y + 2 = 0$.

Solution. The line is $y = -x – 2$. Meeting points: $-x^2 = -x – 2$, i.e. $x^2 – x – 2 = 0$, so $x = -1$ and $x = 2$ — the points $(-1, -1)$ and $(2, -4)$.

The parabola opens downward and the region lies below the $x$-axis, which is where sign errors creep in. The rule does not change: integrate upper minus lower. At $x = 0$ the parabola is at $0$ and the line at $-2$, so the parabola is on top:

$$A = \int_{-1}^{2}\left[-x^2 – (-x – 2)\right]dx = \int_{-1}^{2}\left(x + 2 – x^2\right)dx = \frac92$$

The integrand is identical to Question 6’s: this region is that one reflected in the $x$-axis, so the areas must agree.

Question 15

Find the area bounded by the curve $y = \sqrt{x}$, $x = 2y + 3$ in the first quadrant and $x$-axis.

Solution. The line crosses the $x$-axis at $x = 3$. It meets $y = \sqrt x$, i.e. $x = y^2$ with $y \ge 0$, where $y^2 = 2y + 3$, so $(y – 3)(y + 1) = 0$ and $y = 3$ — the point $(9, 3)$.

Integrating along $x$ would need two pieces: from $0$ to $3$ the region runs from the axis up to the curve, and from $3$ to $9$ from the line up to the curve. Integrating along $y$ needs one, because for every $y \in [0, 3]$ the region runs from the curve on the left ($x = y^2$) to the line on the right ($x = 2y + 3$):

$$A = \int_0^3\left(2y + 3 – y^2\right)dy = \left[y^2 + 3y – \frac{y^3}{3}\right]_0^3 = 9 + 9 – 9$$

Check along $x$. $\int_0^9\sqrt x\,dx – \int_3^9\frac{x – 3}{2}\,dx = 18 – 9 = 9$. Same answer, twice the work.

Common mistakes

• Integrating only the upper branch of a sideways parabola in Question 7. $y^2 = 8x$ is two branches; $\sqrt{8x}$ is one of them. A region symmetric about the $x$-axis needs the factor of 2.
• Taking the meeting points from a sketch. In Question 3 the cubic $x^3 = x + 6$ has to be solved, and the check that its quadratic factor has no real roots is what proves the region has only one boundary point besides the axis.
• Writing lower minus upper when the region is below the axis, as in Question 14. Being below the $x$-axis does not reverse the rule; “upper minus lower” is about the two curves, not about the sign of $y$.
• Splitting Question 15 into two integrals along $x$. It works, but it doubles the arithmetic and doubles the chances of an error. When the boundaries are naturally $x = f(y)$, integrate along $y$.
• Forgetting that a line can cross the axis inside the interval. In Question 10 it does not — but checking that $2y = 5x + 7$ stays positive on $[2, 8]$ is part of the solution, because a crossing would change the integral into two pieces with a sign flip.
• Reading Question 8’s set literally. As printed, $\{(x, 0) : \ldots\}$ is a set of points on the axis. Say what the intended region is before integrating.

Practise next

• Exemplar Long Answer and Objective Q16–34, the rest of the unit — circles against parabolas, triangles by integration, and eleven multiple-choice questions.
• NCERT textbook Exercise 8.1 for area under a single curve, the part of this topic the current textbook still teaches.
• NCERT textbook Miscellaneous Exercise on Application of Integrals, which includes the $y = \sin x$ and $y = |x + 3|$ areas that the Long Answer band builds on.
• Exemplar Integrals Short Answer Q18–34, if evaluating the definite integrals here was the slow part rather than setting them up.