Further Maths Further integration techniques Practice Questions

Free Further Maths Further integration techniques practice questions with full step-by-step worked solutions. Covers arc-length, cartesian-form, integration, parametric-form. Practise exam-style problems and check your method.

arc-lengthcartesian-formintegrationparametric-formpolar-formsurface-of-revolution
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The curve CC has equation y=cosh⁡(x)y=\cosh{\left(x \right)}. Find the exact length of the arc of CC from x=0x=0 to x=1x=1.
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Worked solution

  1. State the arc-length formula for a cartesian curve

    s=∫011+(dydx)2 dxs=\int_{0}^{1}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx

    The limits are the xx-coordinates of the two ends of the arc.

  2. Differentiate the equation of the curve

    dydx=sinh⁡(x)\frac{dy}{dx}=\sinh{\left(x \right)}

    The gradient is what feeds into the arc-length integrand.

  3. Square the gradient and add one

    1+(dydx)2=cosh⁡2(x)1+\left(\frac{dy}{dx}\right)^{2}=\cosh^{2}{\left(x \right)}

    This expression must be a perfect square if the integral is to be exact.

  4. State the exact arc length

    s=sinh⁡(1)s=\sinh{\left(1 \right)}

    This is the exact length of the arc, left in exact form.

Answer
sinh⁡(1)\sinh{\left(1 \right)}
Question 2
2 markseasy
The curve CC has polar equation r=cos⁡(θ)+1r=\cos{\left(\theta \right)} + 1. Which of the following integrals gives the length of the arc of CC from θ=0\theta=0 to θ=π2\theta=\frac{\pi}{2}?
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Worked solution

  1. Recall the polar arc-length formula

    s=∫αβr2+(drdθ)2 dθs=\int_{\alpha}^{\beta}\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}\,d\theta

    Both rr and drdθ\frac{dr}{d\theta} appear, squared, under the root.

  2. Differentiate the polar equation

    drdθ=−sin⁡(θ)\frac{dr}{d\theta}=- \sin{\left(\theta \right)}

    This derivative is the second term under the root.

  3. Substitute into the formula

    s=∫0π22cos⁡(θ)+2 dθs=\int_{0}^{\frac{\pi}{2}}\sqrt{2 \cos{\left(\theta \right)} + 2}\,d\theta

    The limits are the values of θ\theta at the ends of the arc.

  4. Select the correct integral

    ∫0π22cos⁡(θ)+2 dθ\int_{0}^{\frac{\pi}{2}}\sqrt{2 \cos{\left(\theta \right)} + 2}\,d\theta

    This is the polar arc-length integral for the given curve.

Answer
∫0π22cos⁡(θ)+2 dθ\int_{0}^{\frac{\pi}{2}}\sqrt{2 \cos{\left(\theta \right)} + 2}\,d\theta
Question 3
4 marksintermediate
A curve has parametric equations x=2t2x=2 t^{2}, y=t3y=t^{3}, for 0≤t≤10\le t\le 1. Which of the following integrals gives the length of the curve?
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Worked solution

  1. Recall the parametric arc-length formula

    s=∫t1t2(dxdt)2+(dydt)2 dts=\int_{t_{1}}^{t_{2}}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt

    Both parametric derivatives appear, squared, under the root.

  2. Differentiate both parametric equations

    dxdt=4t,dydt=3t2\frac{dx}{dt}=4 t,\qquad\frac{dy}{dt}=3 t^{2}

    These are the two quantities that go under the root.

  3. Substitute into the formula

    s=∫01t9t2+16 dts=\int_{0}^{1}t \sqrt{9 t^{2} + 16}\,dt

    The limits are the parameter values at the ends of the arc.

  4. Reject the option that uses xx and yy themselves

    x2+y2≠(dxdt)2+(dydt)2\sqrt{x^{2}+y^{2}}\neq\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}

    The formula uses the derivatives, not the coordinates.

  5. Reject the option that adds the derivatives

    dxdt+dydt≠dsdt\frac{dx}{dt}+\frac{dy}{dt}\neq\frac{ds}{dt}

    Pythagoras requires the squares to be added, then rooted.

  6. Select the correct integral

    ∫01t9t2+16 dt\int_{0}^{1}t \sqrt{9 t^{2} + 16}\,dt

    This is the parametric arc-length integral for the given curve.

Answer
∫01t9t2+16 dt\int_{0}^{1}t \sqrt{9 t^{2} + 16}\,dt
Question 4
6 markshard
The arc of the curve y=x2y=x^{2} between x=0x=0 and x=1x=1 is rotated through 2π2\pi radians about the yy-axis. Which of the following integrals gives the area of the curved surface generated?
Show worked solution

Worked solution

  1. Recall the surface-area formula

    S=2π∫x dsS=2\pi\int x\,ds

    For rotation about the yy-axis the radius of each band is xx.

  2. Write the element of arc length in terms of xx

    ds=1+(dydx)2 dx=4x2+1 dxds=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx=\sqrt{4 x^{2} + 1}\,dx

    The gradient of the curve is dydx=2x\frac{dy}{dx}=2 x.

  3. Substitute into the formula

    S=2π∫01x4x2+1 dxS=2\pi \int_{0}^{1}x \sqrt{4 x^{2} + 1}\,dx

    The radius and the element of arc length are both written in terms of xx.

  4. Reject the volume formula

    π∫y2 dx is a volume, not a surface area\pi\int y^{2}\,dx\ \text{is a volume, not a surface area}

    A volume of revolution uses discs, not bands of arc.

  5. Reject the option with the wrong radius

    the radius must be measured from the y-axis\text{the radius must be measured from the }y\text{-axis}

    Rotating about the yy-axis makes the radius xx.

  6. Reject the option missing the factor 2π2\pi

    each band has circumference 2πx\text{each band has circumference }2\pi x

    Without the 2π2\pi the answer is only the weighted arc length.

  7. Recall the cartesian arc-length formula

    s=∫ab1+(dydx)2 dxs=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx

    This is the formula quoted in the formula book for a curve given as y=f(x)y=f(x).

  8. Recall where the arc-length formula comes from

    (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2}

    Pythagoras on a small element of the curve gives dsdx=1+(dydx)2\frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}.

  9. Recall the parametric arc-length formula

    s=∫t1t2(dxdt)2+(dydt)2 dts=\int_{t_{1}}^{t_{2}}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt

    Dividing the element (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2} by (δt)2\left(\delta t\right)^{2} gives this form.

  10. Select the correct integral

    2π∫01x4x2+1 dx2\pi \int_{0}^{1}x \sqrt{4 x^{2} + 1}\,dx

    This is the surface-area integral for rotation about the yy-axis.

Answer
2π∫01x4x2+1 dx2\pi \int_{0}^{1}x \sqrt{4 x^{2} + 1}\,dx
Question 5
9 markschallenging
A curve has parametric equations x=6t2x=6 t^{2}, y=4t3y=4 t^{3}, for 0≤t≤10\le t\le 1. The arc is rotated through 2π2\pi radians about the xx-axis. Which of the following integrals gives the area of the curved surface generated?
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Worked solution

  1. Recall the surface-area formula for rotation about the xx-axis

    S=2π∫y dsS=2\pi\int y\,ds

    The radius of each band is the yy-coordinate of the curve.

  2. Write the element of arc length in terms of tt

    ds=(dxdt)2+(dydt)2 dt=12tt2+1 dtds=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt=12 t \sqrt{t^{2} + 1}\,dt

    Here dxdt=12t\frac{dx}{dt}=12 t and dydt=12t2\frac{dy}{dt}=12 t^{2}.

  3. Substitute into the formula

    S=2π∫0148t4t2+1 dtS=2\pi \int_{0}^{1}48 t^{4} \sqrt{t^{2} + 1}\,dt

    Both the radius and the arc element are written in terms of the parameter.

  4. Reject the option with radius xx

    rotation about the x-axis gives radius y\text{rotation about the }x\text{-axis gives radius }y

    The radius xx belongs to rotation about the yy-axis.

  5. Reject the option that uses the cartesian arc element

    1+(dydt)2≠dsdt\sqrt{1+\left(\frac{dy}{dt}\right)^{2}}\neq\frac{ds}{dt}

    In parametric form both derivatives must be squared and added.

  6. Recall the cartesian arc-length formula

    s=∫ab1+(dydx)2 dxs=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx

    This is the formula quoted in the formula book for a curve given as y=f(x)y=f(x).

  7. Recall where the arc-length formula comes from

    (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2}

    Pythagoras on a small element of the curve gives dsdx=1+(dydx)2\frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}.

  8. Recall the parametric arc-length formula

    s=∫t1t2(dxdt)2+(dydt)2 dts=\int_{t_{1}}^{t_{2}}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt

    Dividing the element (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2} by (δt)2\left(\delta t\right)^{2} gives this form.

  9. Recall the polar arc-length formula

    s=∫αβr2+(drdθ)2 dθs=\int_{\alpha}^{\beta}\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}\,d\theta

    It follows from the parametric formula with x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta.

  10. Recall the surface-area formula for rotation about the xx-axis

    S=2π∫y dsS=2\pi\int y\,ds

    Each element of arc sweeps out a thin band of radius yy and width dsds.

  11. Recall the surface-area formula for rotation about the yy-axis

    S=2π∫x dsS=2\pi\int x\,ds

    The radius of the band is now the distance xx from the yy-axis.

  12. Recall the integration-by-parts formula

    ∫abudvdx dx=[uv]ab−∫abvdudx dx\int_{a}^{b} u\frac{dv}{dx}\,dx=\left[uv\right]_{a}^{b}-\int_{a}^{b} v\frac{du}{dx}\,dx

    Integration by parts is what turns InI_{n} into an expression involving a lower index.

  13. Recall the Pythagorean identity for the tangent

    1+tan⁡2x=sec⁡2x1+\tan^{2}x=\sec^{2}x

    This identity is what lets a power of sec⁡x\sec x be split off during a reduction.

  14. Recall the Pythagorean identity

    sin⁡2x+cos⁡2x=1\sin^{2}x+\cos^{2}x=1

    It converts between powers of sin⁡x\sin x and powers of cos⁡x\cos x.

  15. Select the correct integral

    2π∫0148t4t2+1 dt2\pi \int_{0}^{1}48 t^{4} \sqrt{t^{2} + 1}\,dt

    This is the parametric surface-area integral for rotation about the xx-axis.

Answer
2π∫0148t4t2+1 dt2\pi \int_{0}^{1}48 t^{4} \sqrt{t^{2} + 1}\,dt

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