Hard Further Maths Methods in calculus Questions

Challenging, exam-style Further Maths Methods in calculus questions with worked solutions. Stretch yourself on the hardest improper-integrals, integration-by-parts, partial-fractions, integration-by-substitution problems.

improper-integralsintegration-by-partspartial-fractionsintegration-by-substitutionmean-valueinverse-tangent-integral
Further Maths34 questionsStep-by-step solutions
Question 1
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]?
Show worked solution

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}
Question 2
9 markschallenging
Which of the following statements about the improper integral ∫0∞1x+1 dx\int_{0}^{\infty}\frac{1}{x+1}\,dx is correct?
Show worked solution

Worked solution

  1. Write down the integral to be evaluated

    I=∫0∞1x+1 dxI=\int_{0}^{\infty}\frac{1}{x+1}\,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

    ∫0∞1x+1 dx=lim⁡t→∞∫0t1x+1 dx\int_{0}^{\infty}\frac{1}{x+1}\,dx=\lim_{t\to\infty}\int_{0}^{t}\frac{1}{x+1}\,dx

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

  4. Recognise a logarithmic integral

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

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

  5. Write down the antiderivative

    F(x)=ln⁡(x+1)F(x)=\ln{\left(x+1\right)}

    Differentiating FF returns the integrand, as required.

  6. Check the antiderivative by differentiating it

    ddx(ln⁡(x+1))=1x+1\frac{d}{dx}\left(\ln{\left(x+1\right)}\right)=\frac{1}{x+1}

    Differentiating FF must return the integrand exactly.

  7. Apply the limits of integration

    ∫0∞1x+1 dx=[ln⁡(x+1)]0∞\int_{0}^{\infty}\frac{1}{x+1}\,dx=\left[\ln{\left(x+1\right)}\right]_{0}^{\infty}

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

  8. Take the limit of the antiderivative as t→∞t\to\infty

    lim⁡t→∞(ln⁡(t+1))=∞\lim_{t\to\infty}\left(\ln{\left(t+1\right)}\right)=\infty

    This limit is the whole content of the improper integral.

  9. Evaluate the antiderivative at the lower limit

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

    The lower limit is substituted into the antiderivative.

  10. Evaluate the partial integral up to t=10t=10

    ∫0101x+1 dx=2.3979\int_{0}^{10}\frac{1}{x+1}\,dx=2.3979

    The partial integral keeps growing; it does not settle down.

  11. Evaluate the partial integral up to t=1000t=1000

    ∫010001x+1 dx=6.90875\int_{0}^{1000}\frac{1}{x+1}\,dx=6.90875

    The partial integral keeps growing; it does not settle down.

  12. Evaluate the partial integral up to t=1000000t=1000000

    ∫010000001x+1 dx=13.8155\int_{0}^{1000000}\frac{1}{x+1}\,dx=13.8155

    The partial integral keeps growing; it does not settle down.

  13. Compare with the pp-test

    ∫1xpwithp=1\int\frac{1}{x^{p}}\quad\text{with}\quad p=1

    The borderline case p=1p=1 is exactly the one that fails to converge.

  14. Interpret the result

    no finite limit⇒no value can be assigned\text{no finite limit}\Rightarrow\text{no value can be assigned}

    Because the limit is infinite the improper integral has no value.

  15. Conclude that the integral diverges

    ∫0∞1x+1 dx has no finite value\int_{0}^{\infty}\frac{1}{x+1}\,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
9 markschallenging
Which of the following is the value of ∫0∞1x2+6x+13 dx\int_{0}^{\infty}\frac{1}{x^{2}+6x+13}\,dx?
Show worked solution

Worked solution

  1. Write down the integral to be evaluated

    I=∫0∞1x2+6x+13 dxI=\int_{0}^{\infty}\frac{1}{x^{2}+6x+13}\,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

    ∫0∞1x2+6x+13 dx=lim⁡t→∞∫0t1x2+6x+13 dx\int_{0}^{\infty}\frac{1}{x^{2}+6x+13}\,dx=\lim_{t\to\infty}\int_{0}^{t}\frac{1}{x^{2}+6x+13}\,dx

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

  4. Complete the square in the denominator

    x2+6x+13=(x+3)2+4x^{2}+6x+13=\left(x+3\right)^{2}+4

    The denominator is now in the standard form u2+a2u^{2}+a^{2}.

  5. 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.

  6. Identify the constant aa (and the shift)

    a=2,u=x+3a=2,\quad u=x+3

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

  7. Write down the antiderivative

    F(x)=arctan⁡(x2+32)2F(x)=\frac{\arctan{\left(\frac{x}{2}+\frac{3}{2}\right)}}{2}

    Differentiating FF reproduces the integrand.

  8. Check the antiderivative by differentiating it

    ddx(arctan⁡(x2+32)2)=1x2+6x+13\frac{d}{dx}\left(\frac{\arctan{\left(\frac{x}{2}+\frac{3}{2}\right)}}{2}\right)=\frac{1}{x^{2}+6x+13}

    Differentiating FF must return the integrand exactly.

  9. Apply the limits of integration

    ∫0∞1x2+6x+13 dx=[arctan⁡(x2+32)2]0∞\int_{0}^{\infty}\frac{1}{x^{2}+6x+13}\,dx=\left[\frac{\arctan{\left(\frac{x}{2}+\frac{3}{2}\right)}}{2}\right]_{0}^{\infty}

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

  10. Take the limit of the antiderivative as t→∞t\to\infty

    lim⁡t→∞(arctan⁡(t2+32)2)=π4\lim_{t\to\infty}\left(\frac{\arctan{\left(\frac{t}{2}+\frac{3}{2}\right)}}{2}\right)=\frac{\pi}{4}

    This limit is the whole content of the improper integral.

  11. Evaluate the antiderivative at the lower limit

    F(0)=arctan⁡(32)2F\left(0\right)=\frac{\arctan{\left(\frac{3}{2}\right)}}{2}

    The lower limit is substituted into the antiderivative.

  12. Subtract to obtain the value of the integral

    ∫0∞1x2+6x+13 dx=(π4)−(arctan⁡(32)2)=−arctan⁡(32)2+π4\int_{0}^{\infty}\frac{1}{x^{2}+6x+13}\,dx=\left(\frac{\pi}{4}\right)-\left(\frac{\arctan{\left(\frac{3}{2}\right)}}{2}\right)=-\frac{\arctan{\left(\frac{3}{2}\right)}}{2}+\frac{\pi}{4}

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

  13. Check the answer numerically

    I≈0.294001I\approx0.294001

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

  14. Check that the sign of the answer is sensible

    1x2+6x+13≥0 on the interval⇒I>0\frac{1}{x^{2}+6x+13}\ge0\text{ on the interval}\Rightarrow I>0

    A non-negative integrand cannot produce a negative value.

  15. 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.

  16. 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.

  17. Select the option equal to this value

    I=−arctan⁡(32)2+π4I=-\frac{\arctan{\left(\frac{3}{2}\right)}}{2}+\frac{\pi}{4}

    This is the exact value required.

Answer
−arctan⁡(32)2+π4-\frac{\arctan{\left(\frac{3}{2}\right)}}{2}+\frac{\pi}{4}
Question 4
9 markschallenging
Which of the following statements about the improper integral ∫0∞x2e−2x dx\int_{0}^{\infty}x^{2}e^{-2x}\,dx is correct?
Show worked solution

Worked solution

  1. Write down the integral to be evaluated

    I=∫0∞x2e−2x dxI=\int_{0}^{\infty}x^{2}e^{-2x}\,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

    ∫0∞x2e−2x dx=lim⁡t→∞∫0tx2e−2x dx\int_{0}^{\infty}x^{2}e^{-2x}\,dx=\lim_{t\to\infty}\int_{0}^{t}x^{2}e^{-2x}\,dx

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

  4. Use integration by parts

    ∫udvdx dx=uv−∫vdudx dx\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx

    The integrand is a product of a polynomial and an exponential.

  5. Choose uu and dvdx\frac{dv}{dx}

    u=x2,dvdx=e−2x,dudx=2x,v=−e−2x2u=x^{2},\quad\frac{dv}{dx}=e^{-2x},\quad\frac{du}{dx}=2x,\quad v=-\frac{e^{-2x}}{2}

    The polynomial is differentiated and the exponential integrated.

  6. Apply the by-parts formula

    ∫x2e−2x dx=−x2e−2x2−∫−xe−2x dx\int x^{2}e^{-2x}\,dx=-\frac{x^{2}e^{-2x}}{2}-\int -xe^{-2x}\,dx

    The remaining integral has a polynomial of lower degree.

  7. Apply integration by parts a second time

    u=2x,dvdx=e−2x,dudx=2,v=−e−2x2u=2x,\quad\frac{dv}{dx}=e^{-2x},\quad\frac{du}{dx}=2,\quad v=-\frac{e^{-2x}}{2}

    The polynomial is differentiated and the exponential integrated.

  8. Substitute the second by-parts result back in

    ∫2xe−2x dx=−xe−2x−∫−e−2x dx\int 2xe^{-2x}\,dx=-xe^{-2x}-\int -e^{-2x}\,dx

    The remaining integral has a polynomial of lower degree.

  9. Complete the integration

    ∫x2e−2x dx=(−2x2−2x−1)e−2x4+c\int x^{2}e^{-2x}\,dx=\frac{\left(-2x^{2}-2x-1\right)e^{-2x}}{4}+c

    Collecting the pieces gives the antiderivative.

  10. Check the antiderivative by differentiating it

    ddx((−2x2−2x−1)e−2x4)=x2e−2x\frac{d}{dx}\left(\frac{\left(-2x^{2}-2x-1\right)e^{-2x}}{4}\right)=x^{2}e^{-2x}

    Differentiating FF must return the integrand exactly.

  11. Apply the limits of integration

    ∫0∞x2e−2x dx=[(−2x2−2x−1)e−2x4]0∞\int_{0}^{\infty}x^{2}e^{-2x}\,dx=\left[\frac{\left(-2x^{2}-2x-1\right)e^{-2x}}{4}\right]_{0}^{\infty}

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

  12. Take the limit of the antiderivative as t→∞t\to\infty

    lim⁡t→∞((−2t2−2t−1)e−2t4)=0\lim_{t\to\infty}\left(\frac{\left(-2t^{2}-2t-1\right)e^{-2t}}{4}\right)=0

    This limit is the whole content of the improper integral.

  13. Evaluate the antiderivative at the lower limit

    F(0)=−14F\left(0\right)=-\frac{1}{4}

    The lower limit is substituted into the antiderivative.

  14. Subtract to obtain the value of the integral

    ∫0∞x2e−2x dx=(0)−(−14)=14\int_{0}^{\infty}x^{2}e^{-2x}\,dx=\left(0\right)-\left(-\frac{1}{4}\right)=\frac{1}{4}

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

  15. Check the answer numerically

    I≈0.25I\approx0.25

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

  16. Select the statement matching the value found

    I=14I=\frac{1}{4}

    The limit is finite, so the improper integral converges to this value.

Answer
The integral converges to 14\text{The integral converges to }\frac{1}{4}
Question 5
9 markschallenging
Evaluate ∫2∞3x2−x dx\int_{2}^{\infty}\frac{3}{x^{2}-x}\,dx. Give your answer in exact form.
Show worked solution

Worked solution

  1. Write down the integral to be evaluated

    I=∫2∞3x2−x dxI=\int_{2}^{\infty}\frac{3}{x^{2}-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

    ∫2∞3x2−x dx=lim⁡t→∞∫2t3x2−x dx\int_{2}^{\infty}\frac{3}{x^{2}-x}\,dx=\lim_{t\to\infty}\int_{2}^{t}\frac{3}{x^{2}-x}\,dx

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

  4. Factorise the denominator

    x2−x=(x−1)xx^{2}-x=\left(x-1\right)x

    The denominator must be a product of linear factors before splitting it up.

  5. Write the integrand in partial fractions

    3(x−1)x≡Ax−1+Bx\frac{3}{\left(x-1\right)x}\equiv \frac{A}{x-1}+\frac{B}{x}

    Each distinct linear factor contributes one term with a constant numerator.

  6. Multiply through by the denominator

    3≡Ax+B(x−1)3\equiv Ax+B\left(x-1\right)

    Clearing the fractions gives an identity true for every xx.

  7. Substitute x=1x=1 to find AA

    3=A(1)⇒A=33=A\left(1\right)\quad\Rightarrow\quad A=3

    Choosing the root of a factor kills every other term of the identity.

  8. Substitute x=0x=0 to find BB

    3=B(−1)⇒B=−33=B\left(-1\right)\quad\Rightarrow\quad B=-3

    Choosing the root of a factor kills every other term of the identity.

  9. State the partial fraction form

    3x2−x≡3x−1−3x\frac{3}{x^{2}-x}\equiv \frac{3}{x-1}-\frac{3}{x}

    This is the integrand rewritten as a sum of simple fractions.

  10. Integrate each fraction as a logarithm

    ∫3x2−x dx=−3ln⁡(x)+3ln⁡(x−1)+c\int \frac{3}{x^{2}-x}\,dx=-3\ln{\left(x\right)}+3\ln{\left(x-1\right)}+c

    Each term is of the form kx−r\frac{k}{x-r}, whose integral is kln⁡∣x−r∣k\ln\left|x-r\right|.

  11. Check the antiderivative by differentiating it

    ddx(−3ln⁡(x)+3ln⁡(x−1))=3x2−x\frac{d}{dx}\left(-3\ln{\left(x\right)}+3\ln{\left(x-1\right)}\right)=\frac{3}{x^{2}-x}

    Differentiating FF must return the integrand exactly.

  12. Apply the limits of integration

    ∫2∞3x2−x dx=[−3ln⁡(x)+3ln⁡(x−1)]2∞\int_{2}^{\infty}\frac{3}{x^{2}-x}\,dx=\left[-3\ln{\left(x\right)}+3\ln{\left(x-1\right)}\right]_{2}^{\infty}

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

  13. Take the limit of the antiderivative as t→∞t\to\infty

    lim⁡t→∞(−3ln⁡(t)+3ln⁡(t−1))=0\lim_{t\to\infty}\left(-3\ln{\left(t\right)}+3\ln{\left(t-1\right)}\right)=0

    This limit is the whole content of the improper integral.

  14. Evaluate the antiderivative at the lower limit

    F(2)=−ln⁡(8)F\left(2\right)=-\ln{\left(8\right)}

    The lower limit is substituted into the antiderivative.

  15. State the final answer

    I=3ln⁡(2)I=3\ln{\left(2\right)}

    This is the exact value required.

Answer
3ln⁡(2)3\ln{\left(2\right)}

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