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CUET UG Integrals and Applications — practice questions

Explore 38 free Integrals and Applications questions in CUET UG 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 38.

Formulae of elementary geometry cannot find areas enclosed by:Read the answer and worked solution →The area under a curve is built from a large number of thin:Read the answer and worked solution →For a vertical strip of height y and width dx, dA equals:Read the answer and worked solution →The area of one thin strip is called the:Read the answer and worked solution →The total area comes from adding up all the:Read the answer and worked solution →The definite integral is evaluated using which theorem?Read the answer and worked solution →Strips used to find an area may be vertical or:Read the answer and worked solution →If a curve lies below the x-axis, its integral comes out:Read the answer and worked solution →When the integral is negative, the area is taken as its:Read 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 →For the region below the axis A1 is negative while A2 is:Read the answer and worked solution →For a horizontal strip, the elementary area would be:Read the answer and worked solution →Areas are found here for lines and arcs of circles, plus:Read the answer and worked solution →The width of an elementary vertical strip is written as:Read the answer and worked solution →The definite integral was first defined as the limit of:Read the answer and worked solution →The height of a vertical elementary strip is given by:Read the answer and worked solution →An elementary area is located at a position specified by some value of:Read the answer and worked solution →Only the numerical value of an area is taken into consideration, so signs are:Read the answer and worked solution →Finding the area of a circle here uses which curves in standard form?Read 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 →When the integral is negative, the area is taken as its:Read the answer and worked solution →The area of one thin strip is called the:Read the answer and worked solution →Strips used to find an area may be vertical or:Read the answer and worked solution →Only the numerical value of an area is taken into consideration, so signs are:Read the answer and worked solution →The width of an elementary vertical strip is written as:Read the answer and worked solution →For a horizontal strip, the elementary area would be:Read the answer and worked solution →The height of a vertical elementary strip is given by:Read the answer and worked solution →An elementary area is located at a position specified by some value of:Read the answer and worked solution →The area under a curve is built from a large number of thin:Read the answer and worked solution →For a vertical strip of height y and width dx, dA equals:Read the answer and worked solution →

Worked practice questions

Formulae of elementary geometry cannot find areas enclosed by:

  1. Curves
  2. Squares
  3. Circles
  4. Triangl
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Correct answer: A. Curves

1. Simple shapes have known area formulae. 2. A general curved boundary has none. 3. So elementary geometry is inadequate for areas enclosed by curves. 4. The answer is curves. _Source: syllabus Mathematics, Ch 8 'Application of Integrals'_

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The area under a curve is built from a large number of thin:

  1. Points
  2. Strips
  3. Circles
  4. Angles
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Correct answer: B. Strips

1. The region is imagined as sliced up. 2. Each slice is narrow enough to be nearly a rectangle. 3. The area is composed of thin strips. 4. So the answer is strips. _Source: syllabus Mathematics, Ch 8 'Application of Integrals'_

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For a vertical strip of height y and width dx, dA equals:

  1. y dy
  2. x dx
  3. y dx
  4. x dy
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Correct answer: C. y dx

1. A strip is treated as a thin rectangle. 2. Its area is height times width. 3. That gives dA equal to y dx. 4. So the answer is y dx. _Source: syllabus Mathematics, Ch 8 'Application of Integrals'_

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