Mixed Integration

This page is ideal for practising all the skills of integration. You may wish to use this page in preparation for a test on this topic or for the final examinations. The quizzes on this page have been carefully created to take you through all the skills that you need. If you want a more in depth look, then you should go to the individual pages on these topics.


Key Concepts

On this page, you can practise questions on 

  • standard integrals
  • Integration by recognition
  • integration by substitution
  • definite integrals
  • areas between graphs
  • kinematics

Test Yourself

 Here's a quiz that practises the Standard Integrals: \((ax+b)^n , e^{ax}, sin(ax+b), cos(ax+b) \)


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Here is a quiz that practises Integration by Recognition in the form \(\large\int{f'(x)e^{f(x)}{dx}}\)


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Here is a quiz that practises Integration by Recognition in the form \(\large \int f'(x)[f(x)]^{n} {d x}\)


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Here's a quiz that practises Integration by Substitution


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Here's a quiz that practises Definite Integration


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Here's a quiz that practises Areas between Graphs


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Here's a quiz that practises Kinematics


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Here's a quiz that practises Volumes of Revolution


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 Medium

1

The following integral can be performed by using integration by substitution (or U-substitution). Which is the correct substitution?

\(\int { sin^2x \ cosx \ dx } \)

If u = sinx

then \(\frac { du }{ dx } =cosx\)

Integrand becomes \(\int { { u }^{ 2 }du } \)

2

The graph below shows the function f

Evaluate the following definite integral

\(\int _{ -1 }^{ 0 }{ f(x)dx } \) =


definite integral is negative as it is below x axis

3

The graph below shows the function f

\(\int _{ a }^{ 7 }{ f(x)dx } =0\)

Find 2 possible values of a , a1 and a2

\(a_1\) =
\(a_2\) =

You must choose a value for a that makes the definite integral from a to 7 equal 0.

Choose a value of a, so that the area of the region below the x axis must equal the area above the x axis.

    4

    A rocket starts from rest and accelerates such that \(a(t)=3t+t^2\), where t = time

    Find a formula for the velocity of the rocket at time t

    In order to find velocity, we need to integrate the acceleration formula

    \(v(t)=\int { (3t+t^{ 2 })dt } \\ =\frac { 3t^{ 2 } }{ 2 } +\frac { { t }^{ 3 } }{ 3 } +C\)

    The rocket starts from rest, so when t = 0, v = 0

    \(0=\frac { 3(0^{ 2 }) }{ 2 } +\frac { { 0 }^{ 3 } }{ 3 } +C\)

    C = 0

    \(v(t) =\frac { 3t^{ 2 } }{ 2 } +\frac { { t }^{ 3 } }{ 3 } \)

    5

    Work out \(\large\int\frac{2x}{x^2+1} \mathrm{d}x\)

    \(\frac{\mathrm{d}}{\mathrm{d}x}(x^2+1)=2x\)

    Hence integral is in the form \(\int\frac{f'(x)}{f(x)} \mathrm{d}x=\ln |f(x)|+C\)

    \(\int\frac{2x}{x^2+1} \mathrm{d}x=\ln |x^2+1|+C\)


     Hard

    1

    Given that \(\int _{ -2 }^{ 2 }{ f(x)dx=5 } \) and \(\int _{ -2 }^{ 0 }{ f(x)dx=3 } \) work out

    a) \(\int _{ 0 }^{ 2 }{ 2f(x)dx } \) =


    b) \(\int _{ 2 }^{ 4 }{ f(x-2)dx } \) =


    c) \(\int _{ -2 }^{ 2 }{ (f(x)+2)dx } \) =

    a) Consider the same definite integral split into two parts

    \(\int _{ -2 }^{ 0 }{ f(x)dx}+\int _{ 0 }^{ 2 }{ f(x)dx } =\int _{ -2 }^{ 2 }{ f(x)dx } \\3 + 2 = 5\\\int _{ 0 }^{ 2 }{ f(x)dx }=2\)

    b) f(x - 2) translates the graph of f(x) 2 units to the right. This is the same as \(\int _{ 0 }^{ 2 }{ f(x)dx } \)

    c) f(x) + 2 translates the graph 2 units up. It creates a region the same as \(\int _{ -2 }^{ 2 }{ f(x)dx=5 } \) , but with a rectangle below. The area of the rectangle is 8.

    2

    Work out \(\int { \frac { lnx }{ x } dx } \)

    We can use integration by substitution. Let \(u=lnx\)

    \(\int { \frac { lnx }{ x } dx } \)
    \(u=lnx\\ \frac { du }{ dx } =\frac { 1 }{ x } \\ du=\frac { 1 }{ x } dx\)
    \(\int { udu } \)
    \(=\frac { { u }^{ 2 } }{ 2 } +C\)
    \(=\frac { { \left( lnx \right) }^{ 2 } }{ 2 } +C\)
    3

    The diagram below shows the graph of \(y = \frac{2x}{x^2-1}\)

    The shaded region has an area \(lna\)

    Find the value of a

    a =

    Area = \(\int _{ 2 }^{ 4 }{ \frac { 2x }{ x²-1 } dx } \)

    We might recognise that this integral is in the form \(=\int { \frac { f'(x) }{ f(x) } dx=ln|f(x)|+C } \)

    Or, we could use Integration by substitution with the substitution u = x² - 1

    \(\int _{ 2 }^{ 4 }{ \frac { 2x }{ x²-1 } dx } ={ \left[ ln|x²-1| \right] }_{ 2 }^{ 4 }\)

    = ln 15 - ln 3

    \(=ln\frac { 15 }{ 3 } \)

    = ln3

    4

    Work out \(\large\int\frac{1}{x\ln x} \mathrm{d}x\)

    \(\int\frac{1}{x\ln x} \mathrm{d}x=\int\frac{\frac{1}{x}}{\ln x} \mathrm{d}x\)

    \(\frac{\mathrm{d}}{\mathrm{d}x}(\ln x)=\frac{1}{x}\)

    Hence integral is in the form \(\int\frac{f'(x)}{f(x)} \mathrm{d}x=\ln |f(x)|+C\)

    \(\int\frac{\frac{1}{x}}{\ln x} \mathrm{d}x=\ln|\ln x|+C\)

    5

    Work out \(\large \int_0^{1} \frac {x(e^{x^2}+1)}{e^{x^2}+x^2} {d x}\)

    This is an integration by recognition question in the form \(\large \int \frac {f'(x)}{f(x)} {d x}=\ln |f(x)|+C\)

    We can also use integration by substitution using the substitution \(\large u=\cos x\)

    Note that, \(\large \frac{d}{dx}(e^{x^2}+x^2)=2xe^{x^2}+2x=2x(e^{x^2}+1)\)

    So, we have

    \(\large \frac{1}{2}\int_0^{1} \frac {2x(e^{x^2}+1)}{e^{x^2}+x^2} {d x}=\frac{1}{2}[\ln |e^{x^2}+x^2|]_0^{1}\\ \large =\frac{1}{2}\ln([e^1+1]-[e^0+0]) \\ \large=\frac{1}{2}\ln\frac{e+1}{1}\\ \large=\frac{1}{2}\ln(e+1)\)


    MY PROGRESS

    How much of Mixed Integration have you understood?