A-Level Areas and further applications Practice Questions

Free A-Level Areas and further applications practice questions with full step-by-step worked solutions. Covers areas, definite-integral, set-up, identify. Practise exam-style problems and check your method.

areasdefinite-integralset-upidentifyrootsbetween-curves
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Find the area of the region bounded by the curve y=x2y=x^{2}, the xx-axis and the lines x=0x=0 and x=3x=3.
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Worked solution

  1. Write the area as a definite integral

    A=∫03x2 dxA=\int_{0}^{3} x^{2}\,dx

    The area under a curve above the x-axis is a definite integral of y.

  2. Integrate and evaluate between the limits

    A=[x33]03=9A=\left[\frac{x^{3}}{3}\right]_{0}^{3}=9

    Substituting the limits into the antiderivative gives the area.

  3. State the exact area

    A=9A=9

    This is the required area.

Answer
99
Question 2
2 markseasy
Which of the following integrals gives the area of the region bounded by the curve y=3x2y=3 x^{2}, the xx-axis and the lines x=0x=0 and x=2x=2?
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Worked solution

  1. Recall the area under a curve is a definite integral

    A=∫aby dxA=\int_{a}^{b} y\,dx

    The area between a curve and the x-axis is a definite integral.

  2. Identify the integrand and the limits

    y=3x2, a=0, b=2y=3 x^{2},\ a=0,\ b=2

    Read the function and the x-values bounding the region.

  3. Select the correct integral

    ∫023x2 dx\int_{0}^{2} 3 x^{2}\,dx

    This integral gives the required area.

Answer
∫023x2 dx\int_{0}^{2} 3 x^{2}\,dx
Question 3
3 marksintermediate
Which of the following integrals gives the area of the region bounded by the curve y=1xy=\frac{1}{x}, the xx-axis and the lines x=1x=1 and x=4x=4?
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Worked solution

  1. Recall the area under a curve is a definite integral

    A=∫aby dxA=\int_{a}^{b} y\,dx

    The area between a curve and the x-axis is a definite integral.

  2. Identify the integrand and the limits

    y=1x, a=1, b=4y=\frac{1}{x},\ a=1,\ b=4

    Read the function and the x-values bounding the region.

  3. Write the integrand

    y=1xy=\frac{1}{x}

    The integrand is the equation of the curve.

  4. Note the lower limit

    x=1x=1

    The left boundary of the region is the lower limit.

  5. Note the upper limit

    x=4x=4

    The right boundary of the region is the upper limit.

  6. Select the correct integral

    ∫141x dx\int_{1}^{4} \frac{1}{x}\,dx

    This integral gives the required area.

Answer
∫141x dx\int_{1}^{4} \frac{1}{x}\,dx
Question 4
5 markshard
Which of the following integrals gives the area of the region bounded by the curve y=x3+2y=x^{3} + 2, the xx-axis and the lines x=1x=1 and x=3x=3?
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Worked solution

  1. Recall the area under a curve is a definite integral

    A=∫aby dxA=\int_{a}^{b} y\,dx

    The area between a curve and the x-axis is a definite integral.

  2. Identify the integrand and the limits

    y=x3+2, a=1, b=3y=x^{3} + 2,\ a=1,\ b=3

    Read the function and the x-values bounding the region.

  3. Write the integrand

    y=x3+2y=x^{3} + 2

    The integrand is the equation of the curve.

  4. Note the lower limit

    x=1x=1

    The left boundary of the region is the lower limit.

  5. Note the upper limit

    x=3x=3

    The right boundary of the region is the upper limit.

  6. Assemble the definite integral

    ∫13x3+2 dx\int_{1}^{3} x^{3} + 2\,dx

    Combining the integrand and limits gives the required integral.

  7. Reject swapping the limits

    ∫31x3+2 dx (wrong sign)\int_{3}^{1} x^{3} + 2\,dx\ \text{(wrong sign)}

    Swapping the limits reverses the sign, so this is incorrect.

  8. Reject the volume-of-revolution formula

    π∫13y2 dx\pi\int_{1}^{3} y^{2}\,dx

    That formula gives a volume of revolution, not an area.

  9. Reject integrating the gradient function

    ∫133x2 dx\int_{1}^{3} 3 x^{2}\,dx

    Integrating the derivative does not give the area under the curve.

  10. Select the correct integral

    ∫13x3+2 dx\int_{1}^{3} x^{3} + 2\,dx

    This integral gives the required area.

Answer
∫13x3+2 dx\int_{1}^{3} x^{3} + 2\,dx
Question 5
8 markschallenging
Find the area of the region bounded by the curve y=cos⁡(x)y=\cos{\left(x \right)}, the xx-axis and the lines x=0x=0 and x=π3x=\frac{\pi}{3}.
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Worked solution

  1. Write the area as a definite integral

    A=∫0π3cos⁡(x) dxA=\int_{0}^{\frac{\pi}{3}} \cos{\left(x \right)}\,dx

    The area under a curve above the x-axis is a definite integral of y.

  2. Integrate the function to find an antiderivative

    ∫cos⁡(x) dx=sin⁡(x)+c\int \cos{\left(x \right)}\,dx=\sin{\left(x \right)}+c

    Integrating the integrand gives an antiderivative; the constant cancels for a definite integral.

  3. Evaluate the antiderivative at the upper limit

    [sin⁡(x)]x=π3=32\left[\sin{\left(x \right)}\right]_{x=\frac{\pi}{3}}=\frac{\sqrt{3}}{2}

    Substitute the upper limit into the antiderivative.

  4. Evaluate the antiderivative at the lower limit

    [sin⁡(x)]x=0=0\left[\sin{\left(x \right)}\right]_{x=0}=0

    Substitute the lower limit into the antiderivative.

  5. Subtract the lower value from the upper value

    32−(0)=32\frac{\sqrt{3}}{2}-\left(0\right)=\frac{\sqrt{3}}{2}

    The definite integral is the difference of the antiderivative values.

  6. Write the value of the definite integral

    ∫0π3cos⁡(x) dx=32\int_{0}^{\frac{\pi}{3}} \cos{\left(x \right)}\,dx=\frac{\sqrt{3}}{2}

    This is the exact value of the integral.

  7. Recall the fundamental theorem of calculus

    ∫0π3y dx=[F(x)]0π3\int_{0}^{\frac{\pi}{3}} y\,dx=\left[F(x)\right]_{0}^{\frac{\pi}{3}}

    A definite integral equals the change in an antiderivative.

  8. Differentiate the antiderivative to check

    ddx(sin⁡(x))=cos⁡(x)\frac{d}{dx}\left(\sin{\left(x \right)}\right)=\cos{\left(x \right)}

    Differentiating returns the integrand, confirming the antiderivative.

  9. Interpret the integral as an area

    area=32\text{area}=\frac{\sqrt{3}}{2}

    The region lies above the axis, so the area equals the integral.

  10. Express the area as a decimal to 3 s.f.

    A≈0.866A\approx 0.866

    A decimal value shows the size of the area.

  11. Note the constant of integration cancels

    [sin⁡(x)+c]0π3=32\left[\sin{\left(x \right)}+c\right]_{0}^{\frac{\pi}{3}}=\frac{\sqrt{3}}{2}

    Any constant cancels when the limits are substituted.

  12. State the limits of integration

    x=0 to x=π3x=0\ \text{to}\ x=\frac{\pi}{3}

    These limits are the boundaries of the region.

  13. Recall the area formula

    A=∫0π3y dxA=\int_{0}^{\frac{\pi}{3}} y\,dx

    The area is the integral of y with respect to the variable.

  14. Confirm the area is positive

    A=32>0A=\frac{\sqrt{3}}{2}>0

    An area must be positive, which checks the result.

  15. State the exact area

    A=32A=\frac{\sqrt{3}}{2}

    This is the required area.

Answer
32\frac{\sqrt{3}}{2}

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