Definite integration and areas Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Definite integration and areas questions. See exactly how to solve problems on definite-integral, power-rule, linear, constant.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Evaluate 12(3x2)dx\displaystyle \int_{1}^{2} \left( 3 x^{2} \right) \, dx.

Worked solution

  1. Write down the definite integral

    12(3x2)dx\int_{1}^{2} \left( 3 x^{2} \right) \, dx

    Identify the function to integrate and note the lower limit (bottom number) and upper limit (top number).

  2. Integrate using the reverse power rule

    =[x3]12= \left[ x^{3} \right]_{1}^{2}

    Add one to each power and divide by the new power, then keep the limits attached to the square bracket.

  3. Substitute the limits (top minus bottom)

    =(8)(1)= \left(8\right) - \left(1\right)

    Put in the upper limit first, then subtract the value you get from the lower limit.

  4. Work out the value

    =7= 7

    Finish the arithmetic to get the single number.

Answer
77
Question 2
2 markseasy
Evaluate 04(2x)dx\displaystyle \int_{0}^{4} \left( 2 x \right) \, dx.

Worked solution

  1. Write down the definite integral

    04(2x)dx\int_{0}^{4} \left( 2 x \right) \, dx

    Identify the function to integrate and note the lower limit (bottom number) and upper limit (top number).

  2. Integrate using the reverse power rule

    =[x2]04= \left[ x^{2} \right]_{0}^{4}

    Add one to each power and divide by the new power, then keep the limits attached to the square bracket.

  3. Substitute the limits (top minus bottom)

    =(16)(0)= \left(16\right) - \left(0\right)

    Put in the upper limit first, then subtract the value you get from the lower limit.

  4. Work out the value

    =16= 16

    Finish the arithmetic to get the single number.

Answer
1616
Question 3
2 markseasy
Evaluate 12(4x3)dx\displaystyle \int_{1}^{2} \left( 4 x^{3} \right) \, dx.

Worked solution

  1. Write down the definite integral

    12(4x3)dx\int_{1}^{2} \left( 4 x^{3} \right) \, dx

    Identify the function to integrate and note the lower limit (bottom number) and upper limit (top number).

  2. Integrate using the reverse power rule

    =[x4]12= \left[ x^{4} \right]_{1}^{2}

    Add one to each power and divide by the new power, then keep the limits attached to the square bracket.

  3. Substitute the limits (top minus bottom)

    =(16)(1)= \left(16\right) - \left(1\right)

    Put in the upper limit first, then subtract the value you get from the lower limit.

  4. Work out the value

    =15= 15

    Finish the arithmetic to get the single number.

Answer
1515
Question 4
2 markseasy
Evaluate 03(6x+1)dx\displaystyle \int_{0}^{3} \left( 6 x + 1 \right) \, dx.

Worked solution

  1. Write down the definite integral

    03(6x+1)dx\int_{0}^{3} \left( 6 x + 1 \right) \, dx

    Identify the function to integrate and note the lower limit (bottom number) and upper limit (top number).

  2. Integrate using the reverse power rule

    =[3x2+x]03= \left[ 3 x^{2} + x \right]_{0}^{3}

    Add one to each power and divide by the new power, then keep the limits attached to the square bracket.

  3. Substitute the limits (top minus bottom)

    =(30)(0)= \left(30\right) - \left(0\right)

    Put in the upper limit first, then subtract the value you get from the lower limit.

  4. Work out the value

    =30= 30

    Finish the arithmetic to get the single number.

Answer
3030
Question 5
2 markseasy
Evaluate 03(x2+2)dx\displaystyle \int_{0}^{3} \left( x^{2} + 2 \right) \, dx.

Worked solution

  1. Write down the definite integral

    03(x2+2)dx\int_{0}^{3} \left( x^{2} + 2 \right) \, dx

    Identify the function to integrate and note the lower limit (bottom number) and upper limit (top number).

  2. Integrate using the reverse power rule

    =[x33+2x]03= \left[ \frac{x^{3}}{3} + 2 x \right]_{0}^{3}

    Add one to each power and divide by the new power, then keep the limits attached to the square bracket.

  3. Substitute the limits (top minus bottom)

    =(15)(0)= \left(15\right) - \left(0\right)

    Put in the upper limit first, then subtract the value you get from the lower limit.

  4. Work out the value

    =15= 15

    Finish the arithmetic to get the single number.

Answer
1515

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