NCERT Class 12 Mathematics — Application of Integrals, Miscellaneous Exercise on Chapter 8. All 5 questions solved.

Exercise 8.1 kept every region above the $x$-axis. This exercise does not, and that single change is what it tests. Three of the five questions involve a curve that crosses the axis, and in each of them the integral over the whole interval gives the wrong number — because the definite integral counts area below the axis as negative, while area itself is never negative.

Key insight. Area and $\int_a^b y\,dx$ are the same thing only when $y \ge 0$ throughout $[a, b]$. Where the curve crosses the axis, split the interval at the crossing and add the absolute values of the pieces:

$$\text{Area} = \left|\int_a^c y\,dx\right| + \left|\int_c^b y\,dx\right|$$

Questions 2, 3 and 4 are all decided by this. Question 4’s option (A), $-9$, is the value of the unsplit integral — a negative “area”, which the sketch would have ruled out immediately.

Question 1

Find the area under the given curves and given lines:

(i) $y = x^2$, $x = 1$, $x = 2$ and the $x$-axis

(ii) $y = x^4$, $x = 1$, $x = 5$ and the $x$-axis

Solution. Both curves lie entirely above the $x$-axis on the given intervals — $x^2$ and $x^4$ are non-negative everywhere — so no splitting is needed and the integral gives the area directly.

(i)

$$\int_1^2 x^2\,dx = \left[\frac{x^3}{3}\right]_1^2 = \frac83 – \frac13$$

(ii)

$$\int_1^5 x^4\,dx = \left[\frac{x^5}{5}\right]_1^5 = \frac{3125 – 1}{5} = \frac{3124}{5}$$

Question 2

Sketch the graph of $y = |x + 3|$ and evaluate $\displaystyle\int_{-6}^{0}|x+3|\,dx$.

Solution. The graph of $y = |x+3|$ is the graph of $y = |x|$ shifted three units to the left: a V with its vertex at $(-3, 0)$, rising with slope $1$ to the right of the vertex and falling with slope $-1$ to the left.

$$|x+3| = \begin{cases}-(x+3), & x < -3 \\ x+3, & x \ge -3\end{cases}$$

Split the integral at the vertex:

$$\int_{-6}^{-3}-(x+3)\,dx + \int_{-3}^{0}(x+3)\,dx = \frac92 + \frac92$$

The two halves are congruent right-angled triangles of base $3$ and height $3$, so the answer can be read straight off the sketch as $2 \times \tfrac12 \times 3 \times 3$.

Question 3

Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.

Solution. This is the question where the distinction matters most. $\displaystyle\int_0^{2\pi}\sin x\,dx = 0$, because the arch above the axis on $[0, \pi]$ and the arch below it on $[\pi, 2\pi]$ cancel exactly. But the area of the region is not zero.

Split at $x = \pi$, where the curve crosses the axis:

$$\int_0^\pi \sin x\,dx = \Big[-\cos x\Big]_0^\pi = 2$$

$$\left|\int_\pi^{2\pi}\sin x\,dx\right| = \left|\Big[-\cos x\Big]_\pi^{2\pi}\right| = |-2| = 2$$

Choose the correct answer in the following Exercises from 4 to 5.

Question 4

Area bounded by the curve $y = x^3$, the $x$-axis and the ordinates $x = -2$ and $x = 1$ is

Solution. $y = x^3$ is negative for $x < 0$ and positive for $x > 0$, so split at the origin:

$$\left|\int_{-2}^{0}x^3\,dx\right| = \left|\left[\frac{x^4}{4}\right]_{-2}^{0}\right| = |0 – 4| = 4$$

$$\int_0^1 x^3\,dx = \frac14$$

Adding gives $\tfrac{17}{4}$.

Options (A) and (B) are both negative, so neither can be an area. Option (C) is the value of the unsplit integral $\int_{-2}^{1}x^3\,dx = \tfrac14 – 4$ with its sign flipped — right magnitude for the wrong reason, wrong number.

Question 5

The area bounded by the curve $y = x|x|$, the $x$-axis and the ordinates $x = -1$ and $x = 1$ is given by

Solution. As the hint says, unpick the modulus first:

$$y = x|x| = \begin{cases}x^2, & x > 0 \\ -x^2, & x < 0\end{cases}$$

So the curve is the right half of $y = x^2$ joined to the reflection of the left half through the origin — an odd function. Split at $x = 0$:

$$\left|\int_{-1}^{0}\left(-x^2\right)dx\right| + \int_0^1 x^2\,dx = \frac13 + \frac13$$

Option (A) is the value of $\int_{-1}^{1}x|x|\,dx$, which is zero because the function is odd. That is a correct integral and a wrong area — the same trap as question 3, in miniature.

Common mistakes

• Treating $\int_a^b y\,dx$ as the area when the curve crosses the axis. Questions 3, 4 and 5 all punish this, and in question 3 the wrong answer is exactly zero — which should be a visible signal that something is wrong, since the region plainly has area.
• Not sketching. Every crossing point in this exercise is obvious from a sketch and invisible from the algebra. Two minutes on a graph saves the question.
• Splitting at the wrong point. The split goes where $y = 0$, not at the midpoint of the interval or at the origin by default. For $y = |x+3|$ that is $x = -3$, not $x = 0$.
• Reporting a negative area. If your final number is negative you have skipped a modulus. Options (A) and (B) in question 4 exist to catch exactly that.
• Forgetting the units. Areas are in square units; NCERT expects the phrase in the written answer even though the mark is for the number.

Practise next

• Exercise 8.1 — the ellipse and circle areas that this exercise builds on.
• Exercise 7.10 — the modulus integrals of question 31 there use the same splitting argument as questions 2 and 5 here.