Hard A-Level Areas and further applications Questions

Challenging, exam-style A-Level Areas and further applications questions with worked solutions. Stretch yourself on the hardest areas, between-curves, intersection, parametric problems.

areasbetween-curvesintersectionparametricintegrationdefinite-integral
A-Level34 questionsStep-by-step solutions
Question 1
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}.
Show worked solution

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}
Question 2
8 markschallenging
A curve has parametric equations x=t2x=t^{2} and y=t2+1y=t^{2} + 1. Find the area of the region between the curve and the xx-axis for 1≤t≤21\le t\le 2.
Show worked solution

Worked solution

  1. Differentiate x with respect to t

    dxdt=2t\frac{dx}{dt}=2 t

    The parametric area formula needs the rate of change of x.

  2. Set up the parametric area integral

    A=∫12y dxdt dt=∫122t3+2t dtA=\int_{1}^{2} y\,\frac{dx}{dt}\,dt=\int_{1}^{2} 2 t^{3} + 2 t\,dt

    Replace y and dx/dt and integrate with respect to the parameter.

  3. Integrate the function to find an antiderivative

    ∫2t3+2t dt=t42+t2+c\int 2 t^{3} + 2 t\,dt=\frac{t^{4}}{2} + t^{2}+c

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

  4. Evaluate the antiderivative at the upper limit

    [t42+t2]t=2=12\left[\frac{t^{4}}{2} + t^{2}\right]_{t=2}=12

    Substitute the upper limit into the antiderivative.

  5. Evaluate the antiderivative at the lower limit

    [t42+t2]t=1=32\left[\frac{t^{4}}{2} + t^{2}\right]_{t=1}=\frac{3}{2}

    Substitute the lower limit into the antiderivative.

  6. Subtract the lower value from the upper value

    12−(32)=21212-\left(\frac{3}{2}\right)=\frac{21}{2}

    The definite integral is the difference of the antiderivative values.

  7. Write the value of the definite integral

    ∫122t3+2t dt=212\int_{1}^{2} 2 t^{3} + 2 t\,dt=\frac{21}{2}

    This is the exact value of the integral.

  8. Recall the fundamental theorem of calculus

    ∫12y dt=[F(t)]12\int_{1}^{2} y\,dt=\left[F(t)\right]_{1}^{2}

    A definite integral equals the change in an antiderivative.

  9. Differentiate the antiderivative to check

    ddt(t42+t2)=2t3+2t\frac{d}{dt}\left(\frac{t^{4}}{2} + t^{2}\right)=2 t^{3} + 2 t

    Differentiating returns the integrand, confirming the antiderivative.

  10. Interpret the integral as an area

    area=212\text{area}=\frac{21}{2}

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

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

    A≈10.5A\approx 10.5

    A decimal value shows the size of the area.

  12. Note the constant of integration cancels

    [t42+t2+c]12=212\left[\frac{t^{4}}{2} + t^{2}+c\right]_{1}^{2}=\frac{21}{2}

    Any constant cancels when the limits are substituted.

  13. State the limits of integration

    t=1 to t=2t=1\ \text{to}\ t=2

    These limits are the boundaries of the region.

  14. Recall the area formula

    A=∫12y dtA=\int_{1}^{2} y\,dt

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

  15. State the exact area

    A=212A=\frac{21}{2}

    This is the required area.

Answer
212\frac{21}{2}
Question 3
8 markschallenging
Find the area of the region enclosed between the graphs of y=x2y=x^{2} and y=x+6y=x + 6.
Show worked solution

Worked solution

  1. Find where the graphs intersect

    x2=x+6x^{2}=x + 6

    The points of intersection give the limits of integration.

  2. Solve for the x-coordinates of the intersections

    x=−2, 3x=-2,\ 3

    These are the limits a and b of the enclosed region.

  3. Set up the integral of (upper curve minus lower curve)

    A=∫−23(x+6−x2) dx=∫−23−x2+x+6 dxA=\int_{-2}^{3}\left(x + 6-x^{2}\right)\,dx=\int_{-2}^{3} - x^{2} + x + 6\,dx

    Between the limits the upper curve is subtracted from the lower one.

  4. Integrate the function to find an antiderivative

    ∫−x2+x+6 dx=−x33+x22+6x+c\int - x^{2} + x + 6\,dx=- \frac{x^{3}}{3} + \frac{x^{2}}{2} + 6 x+c

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

  5. Evaluate the antiderivative at the upper limit

    [−x33+x22+6x]x=3=272\left[- \frac{x^{3}}{3} + \frac{x^{2}}{2} + 6 x\right]_{x=3}=\frac{27}{2}

    Substitute the upper limit into the antiderivative.

  6. Evaluate the antiderivative at the lower limit

    [−x33+x22+6x]x=−2=−223\left[- \frac{x^{3}}{3} + \frac{x^{2}}{2} + 6 x\right]_{x=-2}=- \frac{22}{3}

    Substitute the lower limit into the antiderivative.

  7. Subtract the lower value from the upper value

    272−(−223)=1256\frac{27}{2}-\left(- \frac{22}{3}\right)=\frac{125}{6}

    The definite integral is the difference of the antiderivative values.

  8. Write the value of the definite integral

    ∫−23−x2+x+6 dx=1256\int_{-2}^{3} - x^{2} + x + 6\,dx=\frac{125}{6}

    This is the exact value of the integral.

  9. Recall the fundamental theorem of calculus

    ∫−23y dx=[F(x)]−23\int_{-2}^{3} y\,dx=\left[F(x)\right]_{-2}^{3}

    A definite integral equals the change in an antiderivative.

  10. Differentiate the antiderivative to check

    ddx(−x33+x22+6x)=−x2+x+6\frac{d}{dx}\left(- \frac{x^{3}}{3} + \frac{x^{2}}{2} + 6 x\right)=- x^{2} + x + 6

    Differentiating returns the integrand, confirming the antiderivative.

  11. Interpret the integral as an area

    area=1256\text{area}=\frac{125}{6}

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

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

    A≈20.83A\approx 20.83

    A decimal value shows the size of the area.

  13. Note the constant of integration cancels

    [−x33+x22+6x+c]−23=1256\left[- \frac{x^{3}}{3} + \frac{x^{2}}{2} + 6 x+c\right]_{-2}^{3}=\frac{125}{6}

    Any constant cancels when the limits are substituted.

  14. State the limits of integration

    x=−2 to x=3x=-2\ \text{to}\ x=3

    These limits are the boundaries of the region.

  15. State the exact area

    A=1256A=\frac{125}{6}

    This is the required area.

Answer
1256\frac{125}{6}
Question 4
8 markschallenging
Find the area of the region bounded by the curve y=1x2y=\frac{1}{x^{2}}, the xx-axis and the lines x=1x=1 and x=3x=3.
Show worked solution

Worked solution

  1. Write the area as a definite integral

    A=∫131x2 dxA=\int_{1}^{3} \frac{1}{x^{2}}\,dx

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

  2. Integrate the function to find an antiderivative

    ∫1x2 dx=−1x+c\int \frac{1}{x^{2}}\,dx=- \frac{1}{x}+c

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

  3. Evaluate the antiderivative at the upper limit

    [−1x]x=3=−13\left[- \frac{1}{x}\right]_{x=3}=- \frac{1}{3}

    Substitute the upper limit into the antiderivative.

  4. Evaluate the antiderivative at the lower limit

    [−1x]x=1=−1\left[- \frac{1}{x}\right]_{x=1}=-1

    Substitute the lower limit into the antiderivative.

  5. Subtract the lower value from the upper value

    −13−(−1)=23- \frac{1}{3}-\left(-1\right)=\frac{2}{3}

    The definite integral is the difference of the antiderivative values.

  6. Write the value of the definite integral

    ∫131x2 dx=23\int_{1}^{3} \frac{1}{x^{2}}\,dx=\frac{2}{3}

    This is the exact value of the integral.

  7. Recall the fundamental theorem of calculus

    ∫13y dx=[F(x)]13\int_{1}^{3} y\,dx=\left[F(x)\right]_{1}^{3}

    A definite integral equals the change in an antiderivative.

  8. Differentiate the antiderivative to check

    ddx(−1x)=1x2\frac{d}{dx}\left(- \frac{1}{x}\right)=\frac{1}{x^{2}}

    Differentiating returns the integrand, confirming the antiderivative.

  9. Interpret the integral as an area

    area=23\text{area}=\frac{2}{3}

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

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

    A≈0.6667A\approx 0.6667

    A decimal value shows the size of the area.

  11. Note the constant of integration cancels

    [−1x+c]13=23\left[- \frac{1}{x}+c\right]_{1}^{3}=\frac{2}{3}

    Any constant cancels when the limits are substituted.

  12. State the limits of integration

    x=1 to x=3x=1\ \text{to}\ x=3

    These limits are the boundaries of the region.

  13. Recall the area formula

    A=∫13y dxA=\int_{1}^{3} y\,dx

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

  14. Confirm the area is positive

    A=23>0A=\frac{2}{3}>0

    An area must be positive, which checks the result.

  15. State the exact area

    A=23A=\frac{2}{3}

    This is the required area.

Answer
23\frac{2}{3}
Question 5
8 markschallenging
Find the area of the region enclosed between the graphs of y=x2+1y=x^{2} + 1 and y=2x+1y=2 x + 1.
Show worked solution

Worked solution

  1. Find where the graphs intersect

    x2+1=2x+1x^{2} + 1=2 x + 1

    The points of intersection give the limits of integration.

  2. Solve for the x-coordinates of the intersections

    x=0, 2x=0,\ 2

    These are the limits a and b of the enclosed region.

  3. Set up the integral of (upper curve minus lower curve)

    A=∫02(2x+1−x2+1) dx=∫02−x2+2x dxA=\int_{0}^{2}\left(2 x + 1-x^{2} + 1\right)\,dx=\int_{0}^{2} - x^{2} + 2 x\,dx

    Between the limits the upper curve is subtracted from the lower one.

  4. Integrate the function to find an antiderivative

    ∫−x2+2x dx=−x33+x2+c\int - x^{2} + 2 x\,dx=- \frac{x^{3}}{3} + x^{2}+c

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

  5. Evaluate the antiderivative at the upper limit

    [−x33+x2]x=2=43\left[- \frac{x^{3}}{3} + x^{2}\right]_{x=2}=\frac{4}{3}

    Substitute the upper limit into the antiderivative.

  6. Evaluate the antiderivative at the lower limit

    [−x33+x2]x=0=0\left[- \frac{x^{3}}{3} + x^{2}\right]_{x=0}=0

    Substitute the lower limit into the antiderivative.

  7. Subtract the lower value from the upper value

    43−(0)=43\frac{4}{3}-\left(0\right)=\frac{4}{3}

    The definite integral is the difference of the antiderivative values.

  8. Write the value of the definite integral

    ∫02−x2+2x dx=43\int_{0}^{2} - x^{2} + 2 x\,dx=\frac{4}{3}

    This is the exact value of the integral.

  9. Recall the fundamental theorem of calculus

    ∫02y dx=[F(x)]02\int_{0}^{2} y\,dx=\left[F(x)\right]_{0}^{2}

    A definite integral equals the change in an antiderivative.

  10. Differentiate the antiderivative to check

    ddx(−x33+x2)=−x2+2x\frac{d}{dx}\left(- \frac{x^{3}}{3} + x^{2}\right)=- x^{2} + 2 x

    Differentiating returns the integrand, confirming the antiderivative.

  11. Interpret the integral as an area

    area=43\text{area}=\frac{4}{3}

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

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

    A≈1.333A\approx 1.333

    A decimal value shows the size of the area.

  13. Note the constant of integration cancels

    [−x33+x2+c]02=43\left[- \frac{x^{3}}{3} + x^{2}+c\right]_{0}^{2}=\frac{4}{3}

    Any constant cancels when the limits are substituted.

  14. State the limits of integration

    x=0 to x=2x=0\ \text{to}\ x=2

    These limits are the boundaries of the region.

  15. State the exact area

    A=43A=\frac{4}{3}

    This is the required area.

Answer
43\frac{4}{3}

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