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Application of Integrals: 5 marks Questions (CBSE Class 12)

6 different 5 marks questions on Application of Integrals from CBSE Class 12 Maths board exams 2026, newest first.

Q335 marksLong AnswerApplication of Integrals CBSE 2026 · 65/4/1

Using integration, find the area of the region enclosed by the curve $y = ∣ x - 6∣$ , the x-axis, and between x = 4 and x = 8.

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Answer: 4 sq. units

$y = ∣ x - 6∣ = 6 - x$ for $x < 6$ and $x - 6$ for $x \geq 6$ .
Area $= \int_{4}^{6} (6 - x) d x + \int_{6}^{8} (x - 6) d x$ .
$= [6 x - \frac{x ^{2}}{2}]_{4}^{6} + [\frac{x ^{2}}{2} - 6 x]_{6}^{8} = (18 - 16) + (- 16 + 18)$ .
$= 2 + 2 = 4$ sq. units.

Q335 marksLong AnswerApplication of Integrals CBSE 2026 · 65/4/2

Using integration, find the area of the region bounded by the curve $y = x ∣ x ∣$ , x-axis, x = $- 2$ and x = 2.

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Answer:

\frac{16}{3}

sq. units

$y = x ∣ x ∣ = x^{2}$ for $x \geq 0$ and $- x^{2}$ for $x < 0$ ; the curve is below the x-axis for $x < 0$ .
Area $= \int_{- 2}^{0} ∣ - x^{2} ∣ d x + \int_{0}^{2} x^{2} d x = \int_{- 2}^{0} x^{2} d x + \int_{0}^{2} x^{2} d x$ .
$= [\frac{x ^{3}}{3}]_{- 2}^{0} + [\frac{x ^{3}}{3}]_{0}^{2} = \frac{8}{3} + \frac{8}{3}$ .
$= \frac{16}{3}$ sq. units.

Q335 marksLong AnswerApplication of Integrals CBSE 2026 · 65/4/3

Using integration, find the area of the region bounded by $y = 5 x + 4$ , $y = 0$ , $x = - 1$ and $x = 1$ .

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Answer:

\frac{41}{5}

sq. units

The line meets the x-axis at $x = - \frac{4}{5}$ ; it is below the axis on $[- 1, - \frac{4}{5}]$ and above on $[- \frac{4}{5}, 1]$ .
Area $= \int_{- 1}^{- 4/5} (5 x + 4) d x + \int_{- 4/5}^{1} (5 x + 4) d x$ .
$\int_{- 1}^{- 4/5} (5 x + 4) d x = [\frac{5 x ^{2}}{2} + 4 x]_{- 1}^{- 4/5} = - \frac{8}{5} + \frac{3}{2} = - \frac{1}{10}$ .
$\int_{- 4/5}^{1} (5 x + 4) d x = \frac{13}{2} - (- \frac{8}{5}) = \frac{81}{10}$ .
Area $= \frac{1}{10} + \frac{81}{10} = \frac{82}{10} = \frac{41}{5}$ sq. units.

Q345 marksLong AnswerApplication of Integrals CBSE 2026 · 65/5/1

Sketch the curve ${(x, y) : 100 x^{2} + 25 y^{2} = 2500}$ and find the area of the region enclosed by it, using integration.

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Answer:

50 π

sq. units

$100 x^{2} + 25 y^{2} = 2500 \Rightarrow \frac{x ^{2}}{25} + \frac{y ^{2}}{100} = 1$ : an ellipse with centre O, x-intercepts $\pm 5$ , y-intercepts $\pm 10$ .
In the first quadrant $y = 2 25 - x^{2}$ ; by symmetry Area $= 4 \int_{0}^{5} 2 25 - x^{2} d x$ .
$\int_{0}^{5} 25 - x^{2} d x = [\frac{x}{2} 25 - x^{2} + \frac{25}{2} sin^{- 1} \frac{x}{5}]_{0}^{5} = \frac{25 π}{4}$ .
Area $= 8 \cdot \frac{25 π}{4} = 50 π$ sq. units.

Q345 marksLong AnswerApplication of Integrals CBSE 2026 · 65/5/2

Sketch the curve described by ${(x, y) : 9 x^{2} + 16 y^{2} = 144}$ and find the area of the region enclosed by it, using integration.

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Answer:

12 π

sq. units

$9 x^{2} + 16 y^{2} = 144 \Rightarrow \frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ : an ellipse with centre O, x-intercepts $\pm 4$ , y-intercepts $\pm 3$ .
In the first quadrant $y = \frac{3}{4} 16 - x^{2}$ ; by symmetry Area $= 4 \int_{0}^{4} \frac{3}{4} 16 - x^{2} d x = 3 \int_{0}^{4} 16 - x^{2} d x$ .
$\int_{0}^{4} 16 - x^{2} d x = [\frac{x}{2} 16 - x^{2} + 8 sin^{- 1} \frac{x}{4}]_{0}^{4} = 4 π$ .
Area $= 3 \cdot 4 π = 12 π$ sq. units.

Q345 marksLong AnswerApplication of Integrals CBSE 2026 · 65/5/3

Sketch the graph defined by ${(x, y) : \frac{x ^{2}}{25} + \frac{y ^{2}}{25} = 1}$ . Find the area of the region of minor segment cut off by the line $x = \frac{5}{2}$ , using integration.

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Answer:

(\frac{25 π}{3} - \frac{25 3}{4})

sq. units

The curve is the circle $x^{2} + y^{2} = 25$ with centre O and radius 5; the line $x = \frac{5}{2}$ cuts it at $(\frac{5}{2}, \pm \frac{5 3}{2})$ .
The minor segment lies between $x = \frac{5}{2}$ and $x = 5$ and is symmetric about the x-axis: Area $= 2 \int_{5/2}^{5} 25 - x^{2} d x$ .
$= 2 [\frac{x}{2} 25 - x^{2} + \frac{25}{2} sin^{- 1} \frac{x}{5}]_{5/2}^{5} = 2 [\frac{25 π}{4} - \frac{25 3}{8} - \frac{25 π}{12}]$ .
$= 2 [\frac{25 π}{6} - \frac{25 3}{8}] = \frac{25 π}{3} - \frac{25 3}{4}$ sq. units.

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