Free physics question bank

Area Under the Curve questions with solutions

Practice Mathematical tools and vector for JEE Main, JEE Advanced, NEET, and Classes 11–12. Attempt each question, then open it for the full solution.

Showing 1–10 of 20 questions

  • MCQeasyMathematical tools and vector

    Find the total area between y = |x| and the x-axis from x = -2 to x = 2.

    Options

    JEE MainsJEE Advanced

    Open question & full solution →
  • MCQmediumMathematical tools and vector

    Find the total area between y = x² - 1 and the x-axis from x = -2 to x = 2.

    Options

    JEE MainsJEE Advanced

    Open question & full solution →
  • MCQmediumMathematical tools and vector

    A particle has velocity v = 4 - 2t m s⁻¹ for 0 ≤ t ≤ 4 s. Find the total distance travelled from the area under the speed-time graph.

    Options

    JEE MainsJEE Advanced

    Open question & full solution →
  • MatchmediumMathematical tools and vector

    Match List I with List II and choose the correct option.

    List 1

    1. y=x and y=x² on [0,1]

    2. y=x² and y=2x on [0,2]

    3. y=x² and y=2-x²

    4. y=x+2 and y=x²

    List 2

    1. 9/2

    2. 8/3

    3. 1/6

    4. 4/3

    Options

    JEE MainsJEE Advanced

    Open question & full solution →
  • MCQmediumMathematical tools and vector

    Find the area enclosed between y = x² and y = 4 - x².

    Options

    JEE MainsJEE Advanced

    Open question & full solution →
  • MCQmediumMathematical tools and vector

    Find the area enclosed between y = x² and y = 2x.

    Options

    JEE MainsJEE Advanced

    Open question & full solution →
  • MatchmediumMathematical tools and vector

    Match List I with List II and choose the correct option.

    List 1

    1. Area under y=x from 0 to 2

    2. Area under y=x² from 0 to 2

    3. Area under y=sinx from 0 to π

    4. Area under y=cosx from 0 to π/2

    List 2

    1. 1 square unit for ∫₀→π/2 cos x dx

    2. 2 square units for area under y = x from 0 to 2

    3. 8/3 square units for area under y = x² from 0 to 2

    4. 2 square units for area under y = sin x from 0 to π

    Options

    JEE MainsJEE Advanced

    Open question & full solution →
  • MCQeasyMathematical tools and vector

    The velocity-time relation of a particle is v = 6 - 2t m s⁻¹ for 0 ≤ t ≤ 3 s. Find its displacement in this interval using area under the v-t curve.

    Options

    JEE MainsJEE Advanced

    Open question & full solution →
  • MCQmediumMathematical tools and vector

    Find the total area between y = x³ - x and the x-axis from x = -1 to x = 1.

    Options

    JEE MainsJEE Advanced

    Open question & full solution →
  • MCQmediumMathematical tools and vector

    Find the area enclosed between y = x² and y = 2 - x².

    Options

    JEE MainsJEE Advanced

    Open question & full solution →