Draw the rough sketch of the curve $y = 20\cos 2x$; (where $\frac{\pi}{6} \le x \le \frac{\pi}{3}$). Using integration, find the area of the region bounded by the curve $y = 20\cos 2x$ from the ordinates $x = \frac{\pi}{6}$ to $x = \frac{\pi}{3}$ and the $x$-axis.

Draw the rough sketch of the curve y = 20cos 2x; (where (π)/(6) ≤ x ≤ (π)/(3)). Using integration, find the ar

Question

y = 20\cos 2x

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✓ Free question

Concept understanding — Area Under Curve

Area Under a Curve

How do you measure the area of a region whose top edge is curved rather than a straight line? For rectangles and triangles we have formulas, but a shape bounded above by $y = f(x)$ has no simple side lengths to plug in. The definite integral is the tool built exactly for this.

The core idea: area as a limit of strips

Take the region under $y = f(x)$ (with $f(x) \ge 0$ ), above the $x$ -axis, between $x = a$ and $x = b$ . Slice it into many thin vertical strips. A strip at position $x$ with tiny width $dx$ is almost a rectangle of height $f(x)$ , so its area is about $f(x)\, dx$ . Add up all the strips and let their width shrink to zero: the sum becomes the definite integral

\text{Area} = \int_a^b f(x)\, dx.

This is why the integral is the area — it is the exact total of infinitely many infinitesimally thin rectangles.

How to compute it

Find an antiderivative $F(x)$ (so $F'(x) = f(x)$ ) and evaluate at the two limits — the Fundamental Theorem of Calculus:

\int_a^b f(x)\, dx = F(b) - F(a).

For example, the area under $y = x^2$ from $0$ to $1$ is $\left[\tfrac{x^3}{3}\right]_0^1 = \tfrac13$ .

Cases you must handle carefully

Watch out

The integral gives signed area. Where the curve dips below the $x$ -axis, $f(x) < 0$ and $\int f\,dx$ comes out negative. For the geometric (positive) area of such a stretch, integrate the absolute value or take the magnitude of that piece: $\displaystyle \text{Area} = \int_a^b |f(x)|\, dx$ .

If a curve crosses the $x$ -axis inside $[a,b]$ , split the integral at each crossing and add the sizes of the parts.

Area with respect to the y-axis

When the region is bounded by a curve $x = g(y)$ and the $y$ -axis between $y = c$ and $y = d$ , slice horizontally instead:

\text{Area} = \int_c^d g(y)\, dy.

Tip

Always sketch the region first. The sketch tells you the correct limits, whether the curve dips below the axis, and whether it is cleaner to integrate in $x$ or in $y$ .

The single big idea: any area with a curved boundary is the sum of infinitely many thin strips, and that sum is precisely a definite integral.

Students searching "Area Under Curve formula and examples" or "Application of Integrals class 12 important questions" will find this the core idea tested throughout NCERT's Application of Integrals chapter, a mainstay of the CBSE Class 12 Maths syllabus and JEE Main/Advanced. Mastering the sign convention for regions below the x-axis is one of the most frequently asked concepts in board and competitive exam papers alike.

Concept: Area Under Curve — the area between a curve and the $x$ -axis is given by $\int_a^b |y| \, dx$ .

Step 1 – Sketch the curve

For $y = 20\cos 2x$ on $x \in [\frac{\pi}{6}, \frac{\pi}{3}]$ :

At $x = \frac{\pi}{6}$ , $y = 20\cos\frac{\pi}{3} = 20 \cdot \frac12 = 10$ (positive).
At $x = \frac{\pi}{4}$ , $y = 20\cos\frac{\pi}{2} = 0$ (crosses $x$ -axis).
At $x = \frac{\pi}{3}$ , $y = 20\cos\frac{2\pi}{3} = 20 \cdot (-\frac12) = -10$ (negative).

So the curve is positive from $\frac{\pi}{6}$ to $\frac{\pi}{4}$ , then negative from $\frac{\pi}{4}$ to $\frac{\pi}{3}$ .

Step 2 – Set up the area integral

Area is the sum of absolute values:

\text{Area} = \int_{\pi/6}^{\pi/4} 20\cos 2x \, dx \;+\; \int_{\pi/4}^{\pi/3} (-20\cos 2x) \, dx

Step 3 – Integrate

Recall $\int \cos 2x \, dx = \frac{\sin 2x}{2}$ .

First integral:

\left[ 20 \cdot \frac{\sin 2x}{2} \right]_{\pi/6}^{\pi/4} = 10\left( \sin\frac{\pi}{2} - \sin\frac{\pi}{3} \right) = 10\left(1 - \frac{\sqrt3}{2}\right)

Second integral:

-10\left[ \sin 2x \right]_{\pi/4}^{\pi/3} = -10\left( \sin\frac{2\pi}{3} - \sin\frac{\pi}{2} \right) = -10\left( \frac{\sqrt3}{2} - 1 \right) = 10\left(1 - \frac{\sqrt3}{2}\right)

Step 4 – Add

Total area = $10\left(1 - \frac{\sqrt3}{2}\right) + 10\left(1 - \frac{\sqrt3}{2}\right) = 20\left(1 - \frac{\sqrt3}{2}\right) = 20 - 10\sqrt3$ .

✓Final answer

\boxed{20 - 10\sqrt{3}}

The area under $y = 20\cos 2x$ from $x = \pi/6$ to $x = \pi/3$ is found by integrating the absolute value of the function, because the curve dips below the $x$ -axis in this interval. The final area is $5$ square units.

Concept and Intuition

When we talk about "area bounded by a curve and the x-axis," we mean the geometric area — always positive. If the curve goes below the x-axis, the ordinary definite integral gives a negative value for that portion. So we must take the absolute value of the function before integrating, or split the integral at the x-intercept.

Here, $y = 20\cos 2x$ is a cosine wave with amplitude 20 and period $\pi$ (since period of $\cos 2x$ is $\pi$ ). Over the interval $\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$ , the angle $2x$ goes from $\frac{\pi}{3}$ to $\frac{2\pi}{3}$ . Cosine is positive at $\frac{\pi}{3}$ ( $\cos 60^\circ = 0.5$ ) but negative at $\frac{2\pi}{3}$ ( $\cos 120^\circ = -0.5$ ). So the curve crosses the x-axis somewhere in between — at $2x = \frac{\pi}{2}$ , i.e., $x = \frac{\pi}{4}$ . That's the key split point.

Step-by-step solution

Find where the curve meets the x-axis Set $y = 0$ :

20\cos 2x = 0 \implies \cos 2x = 0 \implies 2x = \frac{\pi}{2} \implies x = \frac{\pi}{4}

This lies inside $\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$ , so the curve crosses the axis at $x = \pi/4$ .

Determine the sign of $y$ on each side
- For $x \in \left[\frac{\pi}{6}, \frac{\pi}{4}\right)$ : $2x \in \left[\frac{\pi}{3}, \frac{\pi}{2}\right)$ , where cosine is positive. So $y > 0$ .
- For $x \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right]$ : $2x \in \left(\frac{\pi}{2}, \frac{2\pi}{3}\right]$ , where cosine is negative. So $y < 0$ .
Set up the area as a sum of absolute integrals

The geometric area is:

A = \int_{\pi/6}^{\pi/4} 20\cos 2x \, dx \;+\; \int_{\pi/4}^{\pi/3} \bigl(-20\cos 2x\bigr) \, dx

The second integral uses $-y$ because $y$ is negative there.

Integrate $\cos 2x$

Recall: $\int \cos 2x \, dx = \frac{\sin 2x}{2} + C$ .

So $\int 20\cos 2x \, dx = 20 \cdot \frac{\sin 2x}{2} = 10\sin 2x$ .
Evaluate the first integral

\int_{\pi/6}^{\pi/4} 20\cos 2x \, dx = \bigl[10\sin 2x\bigr]_{\pi/6}^{\pi/4}

At $x = \pi/4$ : $\sin(2 \cdot \pi/4) = \sin(\pi/2) = 1$

At $x = \pi/6$ : $\sin(2 \cdot \pi/6) = \sin(\pi/3) = \frac{\sqrt{3}}{2}$

So this part = $10\left(1 - \frac{\sqrt{3}}{2}\right) = 10 - 5\sqrt{3}$ .

Evaluate the second integral

\int_{\pi/4}^{\pi/3} (-20\cos 2x) \, dx = -\bigl[10\sin 2x\bigr]_{\pi/4}^{\pi/3}

At $x = \pi/3$ : $\sin(2\pi/3) = \sin(120^\circ) = \frac{\sqrt{3}}{2}$

At $x = \pi/4$ : $\sin(\pi/2) = 1$

So $-\bigl[10(\frac{\sqrt{3}}{2} - 1)\bigr] = -10\left(\frac{\sqrt{3}}{2} - 1\right) = -5\sqrt{3} + 10 = 10 - 5\sqrt{3}$ .

Add the two parts

A = (10 - 5\sqrt{3}) + (10 - 5\sqrt{3}) = 20 - 10\sqrt{3}

Tip

Notice both integrals gave the same value $10 - 5\sqrt{3}$ . That's because the curve is symmetric about $x = \pi/4$ over this interval — the positive and negative lobes have equal area magnitude. So you could compute one and double it.

Simplify the result

20 - 10\sqrt{3} = 10(2 - \sqrt{3})

Numerically, $\sqrt{3} \approx 1.732$ , so $2 - 1.732 = 0.268$ , and $10 \times 0.268 = 2.68$ . But the exact value is $10(2 - \sqrt{3})$ .

Watch out

A common mistake is to directly integrate $20\cos 2x$ from $\pi/6$ to $\pi/3$ without splitting. That gives $10(\sin\frac{2\pi}{3} - \sin\frac{\pi}{3}) = 10(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}) = 0$ , which is clearly wrong for the geometric area. Always check if the curve crosses the axis within the interval.

✓Final answer

10(2 - \sqrt{3})

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 65/3/11 markMCQ
$\int_{1}^{3} \sqrt{9-y^2}\, dy$
›Reveal solutionSolution
The problem asks for the integral representing the area of a shaded region of a circle. Assuming the shaded region is the quarter circle in the first quadrant of $x^2+y^2=9$ , its area is given by $\boxed{\int_{0}^{3} \sqrt{9-x^2}\, dx}$ .

The core concept here is using definite integrals to calculate the area under a curve. When we have a region bounded by a curve $y=f(x)$ , the x-axis, and vertical lines $x=a$ and $x=b$ , the area is given by $\int_a^b f(x)\, dx$ . Similarly, if the region is bounded by a curve $x=g(y)$ , the y-axis, and horizontal lines $y=c$ and $y=d$ , the area is $\int_c^d g(y)\, dy$ .

The expressions in the options, $\sqrt{9-x^2}$ and $\sqrt{9-y^2}$ , immediately point to the equation of a circle. The general equation of a circle centered at the origin with radius $r$ is $x^2+y^2=r^2$ . Comparing this with $9-x^2$ or $9-y^2$ , we see that $r^2=9$ , which means the radius $r=3$ .

For the upper half of this circle, we can express $y$ as a function of $x$ : $y^2 = 9-x^2 \implies y = \sqrt{9-x^2}$ (taking the positive root for the upper half).

For the right half of this circle, we can express $x$ as a function of $y$ : $x^2 = 9-y^2 \implies x = \sqrt{9-y^2}$ (taking the positive root for the right half).

Since the figure is not provided, we must infer the shaded region from the given options. Options (C) and (D) involve integration from $0$ to $3$ , which is the radius of the circle. This strongly suggests that the shaded region is either a quarter circle or a semi-circle.

Identify the circle's equation and radius:

The terms $\sqrt{9-x^2}$ and $\sqrt{9-y^2}$ indicate that the circle has the equation $x^2+y^2=9$ . This is a circle centered at the origin $(0,0)$ with a radius $r=3$ .

Interpret the shaded region based on options:

Option (C) is $\int_{0}^{3} \sqrt{9-x^2}\, dx$ . This integral represents the area under the curve $y=\sqrt{9-x^2}$ (the upper semi-circle) from $x=0$ to $x=3$ . This region is precisely the quarter circle located in the first quadrant.

Option (D) is $2 \int_{0}^{3} \sqrt{9-x^2}\, dx$ . This would be twice the area of the quarter circle, meaning it represents the area of the entire upper semi-circle (from $x=-3$ to $x=3$ , or by symmetry, $2 \times$ area from $x=0$ to $x=3$ ).

Option (A) is $\int_{1}^{3} \sqrt{9-y^2}\, dy$ . This represents the area under the curve $x=\sqrt{9-y^2}$ (the right semi-circle) from $y=1$ to $y=3$ . This is a specific segment of the quarter circle, not the entire quarter circle.

Option (B) is $2 \int_{1}^{3} \sqrt{9-y^2}\, dy$ . This would be twice the area in (A), representing a horizontal strip of the circle symmetric about the y-axis.

Given that "the shaded region" is singular and the options include a simple quarter-circle integral, the most reasonable assumption for the shaded region, in the absence of a figure, is the quarter circle in the first quadrant. This is a very common setup in such problems.

Set up the integral for the quarter circle:

To find the area of the quarter circle in the first quadrant, we can integrate the function for the upper half of the circle, $y = \sqrt{9-x^2}$ , with respect to $x$ . The x-values for this region range from $0$ to $3$ (the radius).
$A$
Applying this formula, the area of the shaded quarter circle is:

$A = \int_{0}^{3} \sqrt{9-x^2}\, dx$

Compare with the given options:

This integral matches option (C).

Watch out

It is crucial to correctly identify the limits of integration and the function to be integrated. Integrating $\sqrt{9-x^2}$ from $0$ to $3$ gives the area of the quarter circle. Integrating $2\sqrt{9-x^2}$ from $0$ to $3$ would give the area of the entire upper semi-circle. Similarly, integrating $\sqrt{9-y^2}$ from $0$ to $3$ would also give the area of the quarter circle (but with respect to $y$ ).

The correct option is (C).

✓Final answer
$\boxed{\int_{0}^{3} \sqrt{9-x^2}\, dx}$
CBSE 2026Set ANNUAL1 markMCQ
$y=\sin x$
(a) 3
(b) 4
(c) 2
(d) 5
›Reveal solutionSolution

The area under one arch of $y=\sin x$ from $x=0$ to $x=\pi$ is $2$ square units.

Since $\sin x\ge0$ throughout $[0,\pi]$ , the area is simply the definite integral:

$A=\int_0^{\pi}\sin x\,dx=\big[-\cos x\big]_0^{\pi}$

Evaluate:

$A=(-\cos\pi)-(-\cos 0)=-(-1)-(-1)=1+1=2$

✓Final answer
$2$
CBSE 2026Set ANNUAL1 mark
$x^2 + y^2 = a^2$
›Reveal solutionSolution

By symmetry, the area of the full circle is 4 times the area in the first quadrant, which is found by integration.

Area $=4\displaystyle\int_0^a\sqrt{a^2-x^2}\,dx = 4\left[\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}\right]_0^a = 4\left(\dfrac{a^2}{2}\cdot\dfrac{\pi}{2}\right)=\pi a^2$ .

✓Final answer
$=\pi a^2$
CBSE 2025Set 65/4/11 markMCQ
$y = \sqrt{x}$
›Reveal solutionSolution
The region is the area under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ , which is a standard definite integral. The area equals $\frac{16}{3}$ square units, so the correct option is (C).

The problem asks for the area enclosed by the curve $y = \sqrt{x}$ , the vertical lines $x = 0$ and $x = 4$ , and the x-axis. This is a classic "area under a curve" problem — the region is bounded above by the curve, below by the x-axis, and on the sides by two vertical lines. The key idea is that the area between a curve $y = f(x)$ and the x-axis from $x = a$ to $x = b$ is given by the definite integral $\int_a^b f(x) \, dx$ , provided $f(x) \ge 0$ on that interval. Here, $\sqrt{x}$ is non-negative for $x \ge 0$ , so we can directly integrate.

Watch out

A common mistake is to confuse the area under $y = \sqrt{x}$ with the area under $y = x^2$ or to misapply the power rule. Always check the exponent: $\sqrt{x} = x^{1/2}$ , not $x^2$ .

Let’s work through the calculation step by step.

Set up the integral. The region is bounded by $x = 0$ on the left and $x = 4$ on the right. The curve is $y = \sqrt{x}$ , and the lower boundary is the x-axis ( $y = 0$ ). So the area $A$ is:

$A = \int_{0}^{4} \sqrt{x} \, dx$

Rewrite the integrand. Recall that $\sqrt{x} = x^{1/2}$ . This makes the power rule for integration straightforward:

$A = \int_{0}^{4} x^{1/2} \, dx$

Apply the power rule. The power rule states $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$ . Here $n = \frac{1}{2}$ , so $n+1 = \frac{3}{2}$ :

$A = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4} = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4}$

Evaluate at the limits. First, at $x = 4$ : $4^{3/2} = (4^{1/2})^3 = 2^3 = 8$ . So $\frac{2}{3} \times 8 = \frac{16}{3}$ . At $x = 0$ : $0^{3/2} = 0$ , so the lower limit contributes $0$ . Therefore:

$A = \frac{16}{3} - 0 = \frac{16}{3}$
Tip

You can also think of $4^{3/2}$ as $\sqrt{4^3} = \sqrt{64} = 8$ , which is sometimes easier to compute mentally.

The area is $\frac{16}{3}$ square units, which matches option (C).

✓Final answer
$\frac{16}{3}$
CBSE 2025Set ANNUAL1 markMCQ
$y=e^{x}$
(i) $e(e^{2}-1)$
(ii) $e^{3}-1$
(iii) $e(1-e^{2})$
(iv) $e^{2}(1-e)$
›Reveal solutionSolution

Area under $y=e^x$ from $x=1$ to $x=3$ is $\displaystyle\int_1^3 e^x\,dx$ .

Since $y=e^x > 0$ throughout $[1,3]$ , the region between the curve and the $X$ -axis has area:

$A = \int_1^3 e^{x}\,dx = \Big[e^{x}\Big]_1^3 = e^{3} - e^{1} = e(e^{2}-1)$

✓Final answer
$e(e^{2}-1)$
CBSE 2025Set ANNUAL1 markMCQ
$x$
(a) $1$
(b) $0$
(c) $-1$
(d) $2$
›Reveal solutionSolution

The required area is $\int_0^{\pi/2}\cos x\,dx$ .

$\text{Area}=\int_0^{\pi/2}\cos x\,dx=\big[\sin x\big]_0^{\pi/2}=\sin\dfrac{\pi}{2}-\sin 0=1-0=1$

✓Final answer
$1$
CBSE 2025Set ANNUAL1 mark
$x^2+y^2=4$
›Reveal solutionSolution

The full circle $x^2+y^2=4$ has radius $2$ ; the first-quadrant portion is one quarter of the full circle.

Full circle area $=\pi r^2=\pi(2)^2=4\pi$ .

Area in the first quadrant $=\dfrac{1}{4}\times4\pi=\pi$ .

(Equivalently, $\displaystyle\int_0^2\sqrt{4-x^2}\,dx=\left[\dfrac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\dfrac{x}{2}\right]_0^2=2\sin^{-1}(1)=2\cdot\dfrac{\pi}{2}=\pi$ .)

✓Final answer
$\pi$
CBSE 2024Set ANNUAL1 markMCQ
$x^2 + y^2 = 2$
(a) $4\pi$ sq. units
(b) $2\sqrt{2}\pi$ sq. units
(c) $4\pi^2$ sq. units
(d) $2\pi$ sq. units
›Reveal solutionSolution

$2\pi$ sq. units — option (d).

The circle $x^2+y^2=2$ has radius $R=\sqrt2$ (comparing with $x^2+y^2=R^2$ ).

Area of a circle $=\pi R^2 = \pi(\sqrt2)^2 = 2\pi$ sq. units.

✓Final answer
$2\pi$
CBSE 2022Set ANNUAL1 markMCQ
$y=x$
(a) 8 sq. units
(b) 4 sq. units
(c) 2 sq. units
(d) 1 sq. unit
›Reveal solutionSolution

The area under a straight line $y=x$ between two vertical lines is a definite integral, here it also equals the area of a trapezium.

Area $=\displaystyle\int_1^3 x\,dx=\left[\dfrac{x^2}{2}\right]_1^3=\dfrac{9}{2}-\dfrac{1}{2}=\dfrac{8}{2}=4$ sq. units.

✓Final answer
$(b) 4 sq. units.$
CBSE 2022Set ANNUAL1 mark
$Q. The area bounded by the curve y = 2x between x = 0, x = 2 and x-axis is ...$
›Reveal solutionSolution

Integrate y = 2x from x = 0 to x = 2 to get the area under the line.

The curve $y=2x$ is a straight line through the origin. The area bounded by it, the x-axis, and the ordinates $x=0$ and $x=2$ is

$\text{Area} = \displaystyle\int_0^2 2x\,dx = \Big[x^2\Big]_0^2 = 4-0 = 4$ square units.

(This matches the geometric check: it's a right triangle with base 2 and height $y(2)=4$ , area $=\tfrac12\times2\times4=4$ .)

✓Final answer
$=4$
CBSE 2020Set HE8231 mark
$x^2+y^2=4$
›Reveal solutionSolution

The required area is $\pi$ square units — a quarter of the circle $x^2+y^2=4$ .

The circle $x^2+y^2=4$ has centre $(0,0)$ and radius $2$ . In the first quadrant, $y=\sqrt{4-x^2}$ for $0\le x\le2$ . The line $y=2$ touches this circle at the single point $(0,2)$ (it is tangent there, since the circle's topmost point is $(0,2)$ ), so bounding the region additionally by $y=2$ and $x=0$ does not remove or add any area beyond the quarter-disc that already lies between the curve and the two axes.

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

$=\left(\frac{2}{2}\sqrt{0}+2\sin^{-1}(1)\right)-\left(0+2\sin^{-1}(0)\right)=2\cdot\frac{\pi}{2}-0=\pi.$

✓Final answer
$=\pi$
CBSE 2018Set ANNUAL1 markMCQ
$x$
(a) $2$
(b) $-1$
(c) $1$
(d) None of these
›Reveal solutionSolution

$\int_0^{\pi/2}\sin x\,dx=1$ .

The area under $y=\sin x$ from $0$ to $\dfrac{\pi}{2}$ is $\int_0^{\pi/2}\sin x\,dx$ .

$=[-\cos x]_0^{\pi/2}=-\cos\dfrac{\pi}{2}+\cos0=0+1=1$ .

✓Final answer
$1$

Concept and Intuition

Step-by-step solution

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