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Revision:OCR Core 4 - Integration

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Revision:OCR Core 4 - Integration

Contents

2. Integration

Integration by parts

A candidate who has followed the syllabus this far will know of the chain and product rules for differentiation, however no equivalents have been derived in integration. Integration by parts is a technique for the integration of a function that is a product.

Consider the following:

\frac{d}{dx} uv = \frac{du}{dx} v + \frac{dv}{dx} u .

Hence:

uv = \int \left( \left[ \frac{du}{dx} v \right] + \left[ \frac{dv}{dx} u \right] \right) \,dx

Hence:

uv = \int \left( \frac{du}{dx} v \right) \,dx + \int \left( \frac{dv}{dx} u \right) \,dx

Hence:

\int \left( \frac{du}{dx} v \right) \,dx = uv - \int \left( \frac{dv}{dx} u \right) \,dx

Or:

\int \left( \frac{dv}{dx} u \right) \,dx = uv - \int \left( \frac{du}{dx} v \right) \,dx .

Example

1. Calculate \int x \sin{x} \,dx .

u = x

\frac{dv}{dx} = \sin{x}

Hence:

v = - \cos{x}

Hence:

[Unparseable or potentially dangerous latex formula. Error 5 : 827x22].

The idea behind this method is to make sure that the value of \frac{dv}{dx} can be integrated such that the calculation can continue. Note that it is perfectly valid to use the other rearrangement of the rule for integration by parts if this will make the calculation easier.

Definite integrals are much the same as indefinite ones:

\int ^{a} _{b} \left( \frac{dv}{dx} u \right) \,dx = \left[ uv \right] ^{a} _{b} - \int ^{a} _{b} \left( \frac{du}{dx} v \right) \,dx .

Direct substitution

Often expressing "x" in terms of a different variable can make an integral possible to do, this is a substitution method.

The method of integration by substitution is the equivalent of the chain rule in differentiation, and as such can be used to integrate functions of functions, for example:

\int \left( \sqrt{ 1 - x^{2} } \right) \,dx .

The key to this method is choosing an appropriate substitution, and knowing the method.

I = \int f(x) \,dx

Hence:

\frac{dI}{dx} = f(x) .

g(u) = x

\frac{dI}{du} = f(x) \times \frac{dx}{du} = f(g(u)) \times \frac{dx}{du} .

Hence:

I = \int \left( f(g(u)) \times \frac{dx}{du} \right) \,du

Therefore if this integration can be done, it is possible to calculate the value of the original integral.

The answer will be in terms of "u", but this can be converted into "x" using the original function and the inverse of it.

Example

1. Calculate \int \left( \sqrt{ 1 - x^{2} } \right) \,dx .

x = \sin{u}

Hence:

x^{2} = \sin ^{2} {u} .

\frac{dx}{du} = \cos{u}

Hence:

\int \left( \sqrt{ 1 - x^{2} } \right) \,dx = \int \left( \sqrt{ 1 - \sin ^{2} {u} } \right) \left( \cos{u} \right) \,du = \int \cos ^{2} {u} \,du .

\int \cos ^{2} {u} \,du = \frac{1}{2} \int \left( 1 + \cos{2u} \right) \,dx = \frac{u}{2} + \frac{1}{4} \sin{2u} + C

x = \sin{u} \implies u = \sin ^{-1} {x}

Hence:

\frac{u}{2} + \frac{1}{4} \sin{2u} + C = \frac{u}{2} + \frac{1}{2} \sin{u} \sqrt{ 1 - \sin ^{2} {u} } + C = \frac{ \sin ^{-1} {x} }{2} + \frac{1}{2} x \sqrt{1 - x^{2} } + C .

There is a consideration that must be made about the conversion between "x", and "u". The issue is that due to their being a function involved, there must be an inverse (such that "u" can be mapped back onto "x") this implies that this function must be a one-one function, and therefore domain restrictions might have to be made.

Definite integrals

With integration by parts there was no difficulty in stating the rule for the definite integrals, however, the method of integration by substitution is not necessarily as easy.

Consider that the final definite integral is in terms of "u", this means that the limits might not be the same as they were when the integral was stated at first. (If x = u - 2 then the limits of 5 \le x \le 10 must be changed).

The advantage of this is that the integral can be calculated directly in terms of "u" (where possible), and as such there can be some work cut out of the process.

x = g(u) \implies \int ^{a} _{b} f(x) \,dx = \int ^{g ^{-1} (a)} _{g ^{-1} (b)} f(g(u)) \times \frac{dx}{du} \,du .

The conversion of the limits of integration is not too hard to understand, as the candidate is simply changing the limits from some value associated with "x" to the corresponding value associated with "u", this is achieved through the use of the inverse of the function which maps "u" onto "x".

Reverse substitution

In some cases it is possible to split the integrand up into the necessary components such that the method of substitution is already there.

Consider that:

\int \left( f(u) \times \frac{du}{dx} \right) \,dx = \int f(u) \,du .

So, consider this in the notation already used:

u = g(x)

And there is a function "h", such that:

h(u) = h(g(x)) = f(x)

\int \left( f(x) \times \frac{du}{dx} \right) \,dx = \int h(u) \,du .

It is better to practice this than to learn it.

There is a special integral that is to be learned (and can be helpful):

y = \ln | f(x) |

Hence:

\frac{dy}{dx} = \frac{ f^{'} (x) }{ f(x) } .

Hence:

\int \frac{ f^{'} (x) }{ f(x) } \,dx = \ln | f(x) | + C .

Choosing a suitable substitution

Often the most difficult part of the integration by substitution method is choosing the correct substitution that will make the integral easier.

The following table sums up some of the common substitutions:

\begin{array}{r|l} Expression & Substitution \\ (ax + b)^{n} & u = ax + b \\ \sqrt[n]{ax + b} & u^{n} = ax + b \\ a - bx^{2} & x = \sqrt{ \frac{a}{b} } \sin{u} \\ a + bx^{2} & x = \sqrt{ \frac{a}{b} } \tan{u} \\ bx^{2} - a & x = \sqrt{ \frac{a}{b} } \sec{u} \\ e^{x} & u = e^{x} \\ \ln{ \left( ax + b \right) } & ax + b = e^{u} \end{array}

Second, third, and any other substitutions may be used to complete the desired calculation, however at the level of Core 4 it is not likely that there will be any integrals that require more than a single substitution.

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