Using integration, find the area of the region bounded by the triangle whose vertices are (2, -2), (4, 5) and

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(2, -2), (4, 5)

CBSECBSE Class XII Board 2020Subjective· 6mImportance★★★★★

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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 of a triangle can be found by integrating the difference between its upper and lower boundary lines. We will find the equations of the three lines forming the triangle, then set up two definite integrals representing the area between the upper boundary (two line segments) and the lower boundary (one line segment). The calculated area is $\boxed{10 \text{ square units}}$ .

The core idea behind using integration to find the area of a region is that a definite integral $\int_a^b f(x) \, dx$ represents the signed area between the curve $y = f(x)$ and the x-axis, from $x=a$ to $x=b$ . When we need to find the area of a region bounded by multiple curves, we often use the principle of "area between curves," which states that the area $A$ between two curves $y_1 = f(x)$ and $y_2 = g(x)$ from $x=a$ to $x=b$ , where $f(x) \ge g(x)$ over the interval, is given by $\int_a^b (f(x) - g(x)) \, dx$ .

For a triangle defined by its vertices, we can visualize it as a region whose upper boundary is formed by two line segments and whose lower boundary is formed by a single line segment. By finding the equations of these lines and identifying the appropriate limits of integration (the x-coordinates of the vertices), we can set up and evaluate the necessary integrals.

Let the given vertices be $A=(2, -2)$ , $B=(4, 5)$ , and $C=(6, 2)$ .

Identify the lines forming the sides of the triangle.

We need to find the equations of the three lines connecting these vertices. The general formula for a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$ .
- Line AB (connecting $A(2, -2)$ and $B(4, 5)$ ):
  
  Slope $m_{AB} = \frac{5 - (-2)}{4 - 2} = \frac{7}{2}$ .
  
  Equation: $y - (-2) = \frac{7}{2}(x - 2)$
  
  $y + 2 = \frac{7}{2}x - 7$
  
  $y_{AB} = \frac{7}{2}x - 9$
- Line BC (connecting $B(4, 5)$ and $C(6, 2)$ ):
  
  Slope $m_{BC} = \frac{2 - 5}{6 - 4} = \frac{-3}{2}$ .
  
  Equation: $y - 5 = \frac{-3}{2}(x - 4)$
  
  $y - 5 = -\frac{3}{2}x + 6$
  
  $y_{BC} = -\frac{3}{2}x + 11$
- Line AC (connecting $A(2, -2)$ and $C(6, 2)$ ):
  
  Slope $m_{AC} = \frac{2 - (-2)}{6 - 2} = \frac{4}{4} = 1$ .
  
  Equation: $y - (-2) = 1(x - 2)$
  
  $y + 2 = x - 2$
  
  $y_{AC} = x - 4$
Determine the integration strategy.

Observe the x-coordinates of the vertices: $x=2$ , $x=4$ , and $x=6$ .

If we project the triangle onto the x-axis, the region spans from $x=2$ to $x=6$ .

The upper boundary of the triangle is formed by line AB from $x=2$ to $x=4$ , and by line BC from $x=4$ to $x=6$ .

The lower boundary of the triangle is formed by line AC from $x=2$ to $x=6$ .

Therefore, the total area $A$ can be calculated as the sum of two integrals:

$A = \int_2^4 (y_{AB} - y_{AC}) \, dx + \int_4^6 (y_{BC} - y_{AC}) \, dx$

Watch out

A common mistake is to simply integrate each line equation separately and sum/subtract. Remember that $\int f(x) \, dx$ gives the area between $f(x)$ and the x-axis. For the area bounded by the triangle, we need to find the area between the upper boundary and the lower boundary of the triangle itself. This is why we subtract the lower function from the upper function.
Set up and evaluate the first integral (from $x=2$ to $x=4$ ).

In this interval, the upper boundary is $y_{AB}$ and the lower boundary is $y_{AC}$ .

$A_1 = \int_2^4 \left( \left(\frac{7}{2}x - 9\right) - (x - 4) \right) \, dx$

$A_1 = \int_2^4 \left( \frac{7}{2}x - x - 9 + 4 \right) \, dx$

$A_1 = \int_2^4 \left( \frac{5}{2}x - 5 \right) \, dx$

Now, we integrate:

$A_1 = \left[ \frac{5}{2} \cdot \frac{x^2}{2} - 5x \right]_2^4$

$A_1 = \left[ \frac{5}{4}x^2 - 5x \right]_2^4$

$A_1 = \left( \frac{5}{4}(4^2) - 5(4) \right) - \left( \frac{5}{4}(2^2) - 5(2) \right)$

$A_1 = \left( \frac{5}{4}(16) - 20 \right) - \left( \frac{5}{4}(4) - 10 \right)$ …

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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 triangle whose vertices are (2,−2),(4,5)(2, -2), (4, 5)(2,−2),(4,5) and (6,2)(6, 2)(6,2).

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Q.Using integration, find the area of the region bounded by the triangle whose vertices are $(2, -2), (4, 5)$ and $(6, 2)$ .

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