Using integration, find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{4} = 1$ and the lines $x = -2$ and $x = 2$.

> [!TLDR] > The area is found by integrating the upper half of the ellipse from $x=-2$ to $x=2$ and doubling. The result is $4\sqrt{3} + \frac{8\pi}{3}$ square units. The problem asks for the area bounded by an ellipse and two vertical lines. The ellipse is $\frac{x^2}{16} + \frac{y^2}{4} = 1$, which is centered at the origin with semi-major axis $a=4$ along the $x$-axis and semi-minor axis $b=2$ along the $y$-axis. The lines $x=-2$ and $x=2$ are vertical lines that cut through the ellipse symmetrically. The key idea: "area under a curve" means integrating $y$ with respect to $x$ between two $x$-limits. But the ellipse is symmetric about the $x$-axis, so the total area between the two vertical lines is twice the area under the upper half of the ellipse. Let’s work through it. 1. **Solve for $y$ from the ellipse equation.** The ellipse is $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Multiply through by 16: $$x^2 + 4y^2 = 16$$ Then $4y^2 = 16 - x^2$, so $y^2 = \frac{16 - x^2}{4}$. Taking the positive root (upper half): $$y = \sqrt{\frac{16 - x^2}{4}} = \frac{1}{2}\sqrt{16 - x^2}$$ This is the function we integrate. 2. **Set up the integral for the upper half.** The region is bounded between $x = -2$ and $x = 2$. The area of the upper half is: $$A_{\text{upper}} = \int_{-2}^{2} \frac{1}{2}\sqrt{16 - x^2} \, dx$$ The total area (upper + lower) is twice that: $$A = 2 \times \frac{1}{2} \int_{-2}^{2} \sqrt{16 - x^2} \, dx = \int_{-2}^{2} \sqrt{16 - x^2} \, dx$$ 3. **Evaluate the integral.** The integral $\int \sqrt{a^2 - x^2} \, dx$ is a standard form. Here $a = 4$. Recall the formula: > [!FORMULA] > $$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$$ This comes from trigonometric substitution $x = a\sin\theta$, which turns the square root into $a\cos\theta$ and the integral into $\int a^2\cos^2\theta \, d\theta$. Applying with $a=4$: $$\int \

Using integration, find the area of the region bounded by the ellipse (x²)/(16) + (y²)/(4) = 1 and the lines x

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\frac{x^2}{16} + \frac{y^2}{4} = 1

CBSECBSE Class XII Board 2024Subjective· 5mImportance★★★★★

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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: …

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The area is found by integrating the upper half of the ellipse from $x=-2$ to $x=2$ and doubling. The result is $4\sqrt{3} + \frac{8\pi}{3}$ square units.

The problem asks for the area bounded by an ellipse and two vertical lines. The ellipse is $\frac{x^2}{16} + \frac{y^2}{4} = 1$ , which is centered at the origin with semi-major axis $a=4$ along the $x$ -axis and semi-minor axis $b=2$ along the $y$ -axis. The lines $x=-2$ and $x=2$ are vertical lines that cut through the ellipse symmetrically.

The key idea: "area under a curve" means integrating $y$ with respect to $x$ between two $x$ -limits. But the ellipse is symmetric about the $x$ -axis, so the total area between the two vertical lines is twice the area under the upper half of the ellipse.

Let’s work through it.

Solve for $y$ from the ellipse equation. The ellipse is $\frac{x^2}{16} + \frac{y^2}{4} = 1$ . Multiply through by 16:

x^2 + 4y^2 = 16

Then $4y^2 = 16 - x^2$ , so $y^2 = \frac{16 - x^2}{4}$ .

Taking the positive root (upper half):

y = \sqrt{\frac{16 - x^2}{4}} = \frac{1}{2}\sqrt{16 - x^2}

This is the function we integrate.

Set up the integral for the upper half. The region is bounded between $x = -2$ and $x = 2$ . The area of the upper half is:

A_{\text{upper}} = \int_{-2}^{2} \frac{1}{2}\sqrt{16 - x^2} \, dx

The total area (upper + lower) is twice that:

A = 2 \times \frac{1}{2} \int_{-2}^{2} \sqrt{16 - x^2} \, dx = \int_{-2}^{2} \sqrt{16 - x^2} \, dx

Evaluate the integral.

The integral $\int \sqrt{a^2 - x^2} \, dx$ is a standard form. Here $a = 4$ .

Recall the formula:
$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$
Applying with $a=4$ :

\int \sqrt{16 - x^2} \, dx = \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4} + C = \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4} + C

Evaluate from $x=-2$ to $x=2$ . Let $F(x) = \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4}$ . Then:

A = F(2) - F(-2)

First, $F(2)$ :

$\sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$ , so $\frac{2}{2} \cdot 2\sqrt{3} = 2\sqrt{3}$ .

$\sin^{-1}\frac{2}{4} = \sin^{-1}\frac{1}{2} = \frac{\pi}{6}$ , so $8 \cdot \frac{\pi}{6} = \frac{4\pi}{3}$ .

Thus $F(2) = 2\sqrt{3} + \frac{4\pi}{3}$ .

Next, $F(-2)$ : …

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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. …
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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:

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

…

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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.

…

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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. …
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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: …

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CBSE 2025Set ANNUAL1 markMCQ
$x$
(a) $1$
(b) $0$
(c) $-1$
(d) $2$
›Reveal solutionSolution

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

…

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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 $=\dfrac{1}{4}\times4\pi=\pi$ .

…

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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 $x^2+y^2=R^2$ ).

…

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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.

…

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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.

…

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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.

…

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

…

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Q.Using integration, find the area of the region bounded by the ellipse x216+y24=1\frac{x^2}{16} + \frac{y^2}{4} = 116x2​+4y2​=1 and the lines x=−2x = -2x=−2 and x=2x = 2x=2.

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Q.Using integration, find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{4} = 1$ and the lines $x = -2$ and $x = 2$ .

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