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ISC Class 12 Integrals and Applications — practice questions

Explore 97 free Integrals and Applications questions in ISC Class 12 Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Integrals and Applications questions and exercises with answers

Showing questions 1–30 of 97.

Using horizontal strips, the area bounded by the curve $x = g(y)$, the $y$-axis and the lines $y = c$ and $y = d$ is given by which integral?Read the answer and worked solution →The total area enclosed by the circle $x^2 + y^2 = a^2$ isRead the answer and worked solution →The area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ isRead the answer and worked solution →The area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ isRead the answer and worked solution →The area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ isRead the answer and worked solution →The area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ isRead the answer and worked solution →The area of the region bounded by the curve $y^2 = 4x$, the $y$-axis and the line $y = 3$ isRead the answer and worked solution →The area of the region bounded by the line $y = 3x + 2$, the $x$-axis and the ordinates $x = -1$ and $x = 1$ isRead the answer and worked solution →The area bounded by the curve $y = \cos x$, the $x$-axis, between $x = 0$ and $x = 2\pi$ isRead the answer and worked solution →The area bounded by the curve $y = \sin x$, the $x$-axis, between $x = 0$ and $x = 2\pi$ isRead the answer and worked solution →The area under the curve $y = x^2$ between $x = 1$, $x = 2$ and the $x$-axis isRead the answer and worked solution →The value of $\int_{-6}^{0} |x + 3|\,dx$ isRead the answer and worked solution →The area bounded by the curve $y = x^3$, the $x$-axis and the ordinates $x = -2$ and $x = 1$ isRead the answer and worked solution →The area bounded by the curve $y = x|x|$, the $x$-axis and the ordinates $x = -1$ and $x = 1$ isRead the answer and worked solution →Match the Integral (ISC 12)Read the answer and worked solution →Order the Steps: Definite Integral (ISC 12)Read the answer and worked solution →Sort: Differentiate vs Integrate (ISC 12)Read the answer and worked solution →Above or below the axisRead the answer and worked solution →Order the strip argumentRead the answer and worked solution →Place the calculus milestonesRead the answer and worked solution →Type the calculus termRead the answer and worked solution →Who is creditedRead the answer and worked solution →One word is wrongRead the answer and worked solution →Geometry or integrationRead the answer and worked solution →Spot the false statementRead the answer and worked solution →Match each calculus ideaRead the answer and worked solution →Definite or indefinite?Read the answer and worked solution →Tap every use of the derivativeRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →If part of a curve is above the axis and part below, the areas are:Read the answer and worked solution →

Worked practice questions

Using horizontal strips, the area bounded by the curve $x = g(y)$, the $y$-axis and the lines $y = c$ and $y = d$ is given by which integral?

  1. $\int_c^d g(x)\,dx$
  2. $\int_c^d g(y)\,dy$
  3. $\int_c^d y\,dx$
  4. $\int_c^d g'(y)\,dy$
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Correct answer: B. $\int_c^d g(y)\,dy$

1. A horizontal elementary strip has width $x = g(y)$ and thickness $dy$, so $dA = x\,dy$. 2. Summing from $y = c$ to $y = d$ gives $A = \int_c^d x\,dy = \int_c^d g(y)\,dy$. 3. The variable of integration is $y$, ruling out forms written in $x$. _Source: syllabus Mathematics Ch 8 "Application of Integrals", p.2_

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The total area enclosed by the circle $x^2 + y^2 = a^2$ is

  1. $2\pi a$
  2. $4\pi a^2$
  3. $\frac{\pi a^2}{2}$
  4. $\pi a^2$
Show answer and solution

Correct answer: D. $\pi a^2$

1. By symmetry the area is $4$ times the first-quadrant region: $A = 4\int_0^a \sqrt{a^2 - x^2}\,dx$. 2. Using $\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$, evaluate from $0$ to $a$. 3. At $x = a$ the bracket is $\frac{a^2}{2}\sin^{-1}1 = \frac{a^2}{2}\cdot\frac{\pi}{2}$. 4. So $A = 4\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi a^2$. _Source: syllabus Mathematics Ch 8 "Application of Integrals", p.2_

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The area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

  1. $\pi a^2 b$
  2. $\pi a b$
  3. $2\pi a b$
  4. $\frac{\pi a b}{2}$
Show answer and solution

Correct answer: B. $\pi a b$

1. By symmetry $A = 4\int_0^a y\,dx$ where $y = \frac{b}{a}\sqrt{a^2 - x^2}$ in the first quadrant. 2. Thus $A = \frac{4b}{a}\int_0^a \sqrt{a^2 - x^2}\,dx = \frac{4b}{a}\left[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a$. 3. The bracket evaluates to $\frac{a^2}{2}\cdot\frac{\pi}{2}$. 4. So $A = \frac{4b}{a}\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi a b$. (Setting $a = b$ recovers $\pi a^2$, ruling out $\pi a^2 b$.) _Source: syllabus Mathematics Ch 8 "Application of Integrals", p.3_

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