Further Maths Methods in calculus Practice Questions

Free Further Maths Methods in calculus practice questions with full step-by-step worked solutions. Covers improper-integrals, power-rule, exponential-integration, mean-value. Practise exam-style problems and check your method.

improper-integralspower-ruleexponential-integrationmean-valuepolynomial-integrationinverse-tangent-integral
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Evaluate ∫1∞1x2 dx\int_{1}^{\infty}\frac{1}{x^{2}}\,dx.
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Worked solution

  1. Write down the integral to be evaluated

    I=∫1∞1x2 dxI=\int_{1}^{\infty}\frac{1}{x^{2}}\,dx

    Identify the integrand and the limits of integration.

  2. State why the integral is improper

    improper: unbounded interval\text{improper: }\text{unbounded interval}

    This integral is improper because the upper limit is infinite.

  3. State the final answer

    I=1I=1

    This is the exact value required.

Answer
11
Question 2
2 markseasy
Which of the following statements about the improper integral ∫1∞1x dx\int_{1}^{\infty}\frac{1}{x}\,dx is correct?
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Worked solution

  1. Write down the integral to be evaluated

    I=∫1∞1x dxI=\int_{1}^{\infty}\frac{1}{x}\,dx

    Identify the integrand and the limits of integration.

  2. State why the integral is improper

    improper: unbounded interval\text{improper: }\text{unbounded interval}

    This integral is improper because the upper limit is infinite.

  3. Replace the offending limit by tt and take a limit

    ∫1∞1x dx=lim⁡t→∞∫1t1x dx\int_{1}^{\infty}\frac{1}{x}\,dx=\lim_{t\to\infty}\int_{1}^{t}\frac{1}{x}\,dx

    The improper integral is DEFINED as this limit; it is not a substitution.

  4. Conclude that the integral diverges

    ∫1∞1x dx has no finite value\int_{1}^{\infty}\frac{1}{x}\,dx\text{ has no finite value}

    The limit defining the improper integral is infinite, so the integral diverges.

Answer
The integral diverges\text{The integral diverges}
Question 3
4 marksintermediate
Which of the following statements about the improper integral ∫011x dx\int_{0}^{1}\frac{1}{x}\,dx is correct?
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Worked solution

  1. Write down the integral to be evaluated

    I=∫011x dxI=\int_{0}^{1}\frac{1}{x}\,dx

    Identify the integrand and the limits of integration.

  2. State why the integral is improper

    improper: f(x)→∞ as x→0\text{improper: }f(x)\to\infty\text{ as }x\to0

    This integral is improper because the integrand is undefined at x=0x=0.

  3. Replace the offending limit by tt and take a limit

    ∫011x dx=lim⁡t→0+∫t11x dx\int_{0}^{1}\frac{1}{x}\,dx=\lim_{t\to0^{+}}\int_{t}^{1}\frac{1}{x}\,dx

    The improper integral is DEFINED as this limit; it is not a substitution.

  4. Recognise a logarithmic integral

    ∫1x dx=1ln⁡∣x∣+c\int\frac{1}{x}\,dx=1\ln\left|x\right|+c

    The numerator is a constant multiple of the derivative of the denominator.

  5. Write down the antiderivative

    F(x)=ln⁡(x)F(x)=\ln{\left(x\right)}

    Differentiating FF returns the integrand, as required.

  6. Conclude that the integral diverges

    ∫011x dx has no finite value\int_{0}^{1}\frac{1}{x}\,dx\text{ has no finite value}

    The limit defining the improper integral is infinite, so the integral diverges.

Answer
The integral diverges\text{The integral diverges}
Question 4
6 markshard
Which of the following is the mean value of the function f(x)=x2f(x)=x^{2} over the interval [−2,4]\left[-2,4\right]?
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Worked solution

  1. Quote the formula for the mean value of a function

    fˉ=1b−a∫abf(x) dx=16∫−24x2 dx\bar{f}=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx=\frac{1}{6}\int_{-2}^{4}x^{2}\,dx

    The mean value is the area under the curve divided by the width of the interval, here b−a=6b-a=6.

  2. Integrate term by term

    ∫(x2) dx=x33+c\int\left(x^{2}\right)\,dx=\frac{x^{3}}{3}+c

    Each power of xx is integrated separately using the power rule.

  3. Check the antiderivative by differentiating it

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

    Differentiating FF must return the integrand exactly.

  4. Apply the limits of integration

    ∫−24x2 dx=[x33]−24\int_{-2}^{4}x^{2}\,dx=\left[\frac{x^{3}}{3}\right]_{-2}^{4}

    The value is the antiderivative at the top minus the antiderivative at the bottom, with a limit taken at any improper endpoint.

  5. Evaluate the antiderivative at the upper limit

    F(4)=643F\left(4\right)=\frac{64}{3}

    The upper limit is substituted into the antiderivative.

  6. Evaluate the antiderivative at the lower limit

    F(−2)=−83F\left(-2\right)=-\frac{8}{3}

    The lower limit is substituted into the antiderivative.

  7. Subtract to obtain the value of the integral

    ∫−24x2 dx=(643)−(−83)=24\int_{-2}^{4}x^{2}\,dx=\left(\frac{64}{3}\right)-\left(-\frac{8}{3}\right)=24

    This is the exact value of the (possibly improper) integral.

  8. Divide by the width of the interval

    fˉ=16×24=4\bar{f}=\frac{1}{6}\times 24=4

    The width of [−2,4]\left[-2,4\right] is 66.

  9. Check the answer numerically

    fˉ≈4.0\bar{f}\approx4.0

    Numerical quadrature agrees with the exact value to six significant figures.

  10. Check that the sign of the answer is sensible

    x2≥0 on the interval⇒fˉ>0x^{2}\ge0\text{ on the interval}\Rightarrow \bar{f}>0

    A non-negative integrand cannot produce a negative value.

  11. Select the option equal to this value

    fˉ=4\bar{f}=4

    This is the exact value required.

Answer
44
Question 5
9 markschallenging
Which of the following is the mean value of the function f(x)=1x2+4f(x)=\frac{1}{x^{2}+4} over the interval [0,2]\left[0,2\right]?
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Worked solution

  1. Quote the formula for the mean value of a function

    fˉ=1b−a∫abf(x) dx=12∫021x2+4 dx\bar{f}=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx=\frac{1}{2}\int_{0}^{2}\frac{1}{x^{2}+4}\,dx

    The mean value is the area under the curve divided by the width of the interval, here b−a=2b-a=2.

  2. Quote the standard inverse-tangent integral

    ∫1a2+u2 du=1aarctan⁡(ua)+c\int\frac{1}{a^{2}+u^{2}}\,du=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+c

    This is the standard result from the formula booklet.

  3. Identify the constant aa

    a=2,u=xa=2,\quad u=x

    Matching the denominator with a2+u2a^{2}+u^{2} fixes aa.

  4. Write down the antiderivative

    F(x)=arctan⁡(x2)2F(x)=\frac{\arctan{\left(\frac{x}{2}\right)}}{2}

    Differentiating FF reproduces the integrand.

  5. Check the antiderivative by differentiating it

    ddx(arctan⁡(x2)2)=1x2+4\frac{d}{dx}\left(\frac{\arctan{\left(\frac{x}{2}\right)}}{2}\right)=\frac{1}{x^{2}+4}

    Differentiating FF must return the integrand exactly.

  6. Apply the limits of integration

    ∫021x2+4 dx=[arctan⁡(x2)2]02\int_{0}^{2}\frac{1}{x^{2}+4}\,dx=\left[\frac{\arctan{\left(\frac{x}{2}\right)}}{2}\right]_{0}^{2}

    The value is the antiderivative at the top minus the antiderivative at the bottom, with a limit taken at any improper endpoint.

  7. Evaluate the antiderivative at the upper limit

    F(2)=π8F\left(2\right)=\frac{\pi}{8}

    The upper limit is substituted into the antiderivative.

  8. Evaluate the antiderivative at the lower limit

    F(0)=0F\left(0\right)=0

    The lower limit is substituted into the antiderivative.

  9. Subtract to obtain the value of the integral

    ∫021x2+4 dx=(π8)−(0)=π8\int_{0}^{2}\frac{1}{x^{2}+4}\,dx=\left(\frac{\pi}{8}\right)-\left(0\right)=\frac{\pi}{8}

    This is the exact value of the (possibly improper) integral.

  10. Divide by the width of the interval

    fˉ=12×π8=π16\bar{f}=\frac{1}{2}\times \frac{\pi}{8}=\frac{\pi}{16}

    The width of [0,2]\left[0,2\right] is 22.

  11. Check the answer numerically

    fˉ≈0.19635\bar{f}\approx0.19635

    Numerical quadrature agrees with the exact value to six significant figures.

  12. Check that the sign of the answer is sensible

    1x2+4≥0 on the interval⇒fˉ>0\frac{1}{x^{2}+4}\ge0\text{ on the interval}\Rightarrow \bar{f}>0

    A non-negative integrand cannot produce a negative value.

  13. Recall the power rule for integration

    ∫xn dx=xn+1n+1+c,n≠−1\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+c,\quad n\neq-1

    Every power except x−1x^{-1} is integrated by raising the index by one.

  14. Recall the definition of an improper integral with an infinite limit

    ∫a∞f(x) dx=lim⁡t→∞∫atf(x) dx\int_{a}^{\infty}f(x)\,dx=\lim_{t\to\infty}\int_{a}^{t}f(x)\,dx

    ∞\infty is not a number, so the upper limit must be replaced by tt and a limit taken.

  15. Recall the definition of an improper integral at a singular endpoint

    ∫abf(x) dx=lim⁡t→a+∫tbf(x) dx\int_{a}^{b}f(x)\,dx=\lim_{t\to a^{+}}\int_{t}^{b}f(x)\,dx

    If ff is undefined at an endpoint the limit is taken from inside the interval.

  16. Select the option equal to this value

    fˉ=π16\bar{f}=\frac{\pi}{16}

    This is the exact value required.

Answer
π16\frac{\pi}{16}

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