07 1antidiff

Antidifferentiation (integration)

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  • This is the reverse of differentiation.
  • Given a gradient function, it allows us to find the original function.

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Recall

… … $\dfrac{d}{dx} \Big( f(x) \Big)$ means differentiate $f(x)$ with respect to x

… … $\dfrac{d}{dx} \Big( f(x) \Big) = f'(x)$
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… … so $f(x)$ is the antiderivative of $f'(x)$
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… … $f(x) = \displaystyle{ \int{ f'(x) } \; dx + c }$
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Notation

  • $\displaystyle{ \int }$ means antidifferentiate (or integrate)

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  • dx means the integration is with respect to the variable x

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  • $\displaystyle{\int } \;\; dx$ … … the two symbols together act as brackets
    • everything inside is integrated

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  • The dx symbol MUST be included every time!!

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  • Sometimes we use $F(x)$ as the antiderivative of $f(x)$

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  • $F(x) + c = \displaystyle{ \int{ f(x) } \; dx } + c$

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Constant of Antidifferentiation (c)

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  • Remember that when we differentiate, any constant term is lost.

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  • So when we antidifferentiate we cannot know the value of the constant term
    • (unless given extra information).
  • Therefore we indicate the unknown constant term with:
    • the constant of antidifferentiation (usually c).

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  • You should (almost) always write the +c when antidifferentiating.

There are only two circumstances when you don't need to write +c

  • 1 … When the question asks you to find an antiderivative.
    • In that case, $c = 0$ is an acceptable solution so
    • you can write the antiderivative without the +c.
  • 2 … When calculating Definite Integrals,
    • the c term cancels out during the calculations so it is not needed.

* It would not be wrong to write +c in these questions so if in doubt, write +c.
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Properties of Integrals

… … $\displaystyle{ \int{ f(x) + g(x) } \; dx} = \displaystyle{ \int{ f(x) } \; dx + \int{ g(x) } \; dx}$
… … … for any functions, f and g

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… … $\displaystyle{ \int{ kf(x) } \; dx} = \displaystyle{ k \int{ f(x) } \; dx}$
… … … for any constant, k

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Polynomials

… … $\displaystyle{ \int {ax^n} \; dx} = \dfrac {ax^{n+1}}{n+1} + c, \quad n \ne -1$
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Example 1

Find $\displaystyle{ \int } x^3 + 4x^2 - x + 3 \, dx$

Solution

… … $\displaystyle{ \int } x^3 + 4x^2 - x + 3 \, dx$

… … … $= \dfrac {x^4}{4} + \dfrac {4x^3}{3} - \dfrac{x^2}{2} + 3x + c$

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Example 2

Find $\displaystyle{\int} \, \dfrac {1}{x^3} + \dfrac {4}{x^2} \, dx$

Solution

… … $\displaystyle{\int} \, \dfrac {1}{x^3} + \dfrac {4}{x^2} \, dx$

… … … $= \displaystyle{ \int} {x^{-3} + 4x^{-2} }dx$

… … … $= \dfrac {x^{-2}}{-2} + \dfrac {4x^{-1}}{-1} + c$

… … … $= -\dfrac {1}{2x^2} - \dfrac {4}{x} + c$

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Example 3

Find $\displaystyle{ \int } \, \sqrt{x} \, dx$

Solution

… … $\displaystyle{ \int } \, \sqrt{x} \, dx$

… … … $= \displaystyle{ \int } x^{ \frac{1}{2} } \, dx$

… … … $= \dfrac {x^{ \frac{3}{2}}} {\frac{3}{2}} + c$

… … … $= \dfrac {2 \sqrt{x^3}} {3} + c$

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Example 4

Find the equation of the curve, $g(x)$,

… … given that $g(1) = 2$ and

… … $g'(x) = 3 \sqrt{x} + 2$

Solution:

… … $g'(x) = 3 \sqrt{x} + 2$

… … … $= 3x^{\frac {1}{2}} + 2$
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… … $g(x) = \displaystyle{ \int{ 3x^{\frac {1}{2}} + 2 }\, dx }$

… … … $= \dfrac {3x^{ \frac{3}{2}} }{ \frac{3}{2} } + 2x + c$

… … … $= \dfrac {2 \times 3x^{ \frac{3}{2}} }{3} + 2x + c$

… … … $= 2x^{ \frac{3}{2}} + 2x + c$
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… … We are given the point $g(1) = 2$

… … … $g(1) = 2(1)^{ \frac{3}{2}} + 2(1) + c = 2$

… … … $2 + 2 + c = 2$

… … … $c = -2$
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… … Hence

… … … $g(x) = 2x^{ \frac{3}{2}} + 2x - 2$

… … Or

… … … $g(x) = 2\sqrt{x^3} +2x - 2$

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Integration of (ax + b)n where $n \neq -1$

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… … Recall that by the chain rule for differentiation:

… … … $\dfrac{d}{dx} \big( ax+b \big)^{n+1} = a(n+1)(ax+b)^n$

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… … Therefore, the reverse is true

… … … $\displaystyle{ \int {a(n+1)(ax+b)^n} \; dx} = \big( ax+b \big)^{n+1} + c$
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… … This can be simplified by moving the constants out of the integral:

… … … $a(n+1) \displaystyle { \int {(ax+b)^n}\; dx} = \big( ax+b \big)^{n+1} + c$
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… … Therefore:

… … … $\displaystyle{ \int {(ax+b)^n}\; dx} = \dfrac{ \big( ax+b \big)^{n+1}}{a(n+1)} + c_1$
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Note: This only works for a linear term raised to a power.
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Note: For Specialist students, it is preferred that you use Integration by Substitution rather than this rule.

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Example 1

Find $\displaystyle{ \int 7(5x-2)^3 \; dx }$

Solution

… … $\displaystyle{ \int 7(5x-2)^3 \; dx }$

… … … $= 7 \displaystyle{ \int (5x-2)^3 \; dx }$

… … … $= \dfrac{7(5x-2)^4}{5\times 4} + c$

… … … $= \dfrac{7(5x-2)^4}{20} + c$

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Integration into Natural Logarithms

… … Since: $\dfrac {d}{dx} \Big( \log_e (x) \Big) = \dfrac {1}{x}$
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… … Then: $\displaystyle{\int} \dfrac{1}{x} \; dx = \log_e (x) + c, \quad \text { where } \; x > 0$ …. (for Methods Students)

… … Then: $\displaystyle{\int} \dfrac{1}{x} \; dx = \log_e |x| + c, \quad \text { where } \; x \neq 0$ …. (for Specialist Students)

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… … Also since: $\dfrac {d}{dx} \Big( \log_e \big( g(x) \big) \Big) = \dfrac {g'(x)}{g(x)}$
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… … Then: $\displaystyle{\int} \dfrac{g'(x)}{g(x)} \; dx = \log_e \big( g(x) \big) + c, \quad \text { where } \; g(x) > 0$ …. (for Methods Students)

… … Then: $\displaystyle{\int} \dfrac{g'(x)}{g(x)} \; dx = \log_e \big| g(x) \big| + c, \quad \text { where } \; g(x) \neq 0$ …. (for Specialist Students)

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Example 1

Find $\displaystyle{\int} \dfrac{5}{2x+3} dx \quad \text{ for } x > -\dfrac{3}{2}$

Solution

… … Adjust numerator so that it is the derivative of the denominator

… … $\displaystyle{\int} \dfrac{5}{2x+3} dx$

… … … $= 5\displaystyle{\int} \dfrac{1}{2x+3} dx$

… … … $= \dfrac{5}{2} \displaystyle{\int} \dfrac{2}{2x+3} dx$

… … … $= \dfrac{5}{2} \log_e (2x+3) + c, \; \; x > -\dfrac {3}{2}$
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Example 2

Find $\displaystyle{\int} \dfrac{6x+5}{x^2} dx \quad \text{ for } x \neq 0$

Solution

… … Separate into two fractions, simplify, then integrate

… … $\displaystyle{\int} \dfrac{6x+5}{x^2} dx$

… … … $= \displaystyle{\int} \dfrac{6x}{x^2} +\dfrac {5}{x^2} dx$

… … … $= \displaystyle{\int} \dfrac{6}{x} +\dfrac {5}{x^2} dx$

… … … $= \displaystyle{\int} \dfrac{6}{x} +5x^{-2} dx$

… … … $= 6\log_e |x| - 5x^{-1} + c$

… … … $= 6\log_e |x| - \dfrac {5}{x} + c, \; \; x \neq 0$

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Using the calculator to find integrals.

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