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Fundamental theorem
of calculus $${\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$$

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Definite Integration

$${\int }_a^{b}f\left(x\right)dx={\left[F\right(x\left)\right]}_a^b=F\left(b\right)-F\left(a\right)$$

where

$$F^{\text{'}}\left(x\right)=f\left(x\right)$$

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Useful Identities

$${\int }_a^{b}kf\left(x\right)dx=k{\int }_a^{b}f\left(x\right)dx$$
$${\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx$$

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Questions

1. $${\int }_1^{5}5x^2-x+2dx$$

Questions from mmerevise.co.uk

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Integration by Parts

Prerequesite Materials:

Product Rule for Differentiation

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Integration by parts uses the product rule for differentiation rearranged to find the integral of a function which is of the form $\int f(x)g(x)dx$. This is given in the formula book as:

$$\int u\frac{dv}{dx}dx=uv-\int \frac{du}{dx}vdx$$

• where
• $u=f(x)$;
• and
• $\frac{dv}{dx}=g(x)$.

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Proof:

Take the product rule for differentiation:

$$\frac{d(uv)}{dx}=\frac{du}{dx}v+u\frac{dv}{dx}$$

Integrate this with respect to $x$,

$$uv=\int \frac{du}{dx}vdx+\int u\frac{dv}{dx}dx$$

Rearrange

$$\int u\frac{dv}{dx}dx=uv-\int \frac{du}{dx}vdx$$

For a given product integrand

$$\int f(x)g(x)dx$$

To use integration by parts on this integral, let $u$ and $v$ equal one or the other factor.

Guidance on how to select correctly:

Let $u$ equal the factor which is a polynomial unless one of the factors is $lnh\left(x\right)$;

Let $\frac{dv}{dx}$ equal the other factor;

If the above is not the situation, then try one way and if it fails then try the other.

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An alternative View
(from www.mathsisfun.com)

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Questions:

1. • $$I=\int {\mathrm{e}}^x(x^2-7)dx$$

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     Solution

   • Using integration by parts, let ...

     $$u=x^2-7\text{\hspace{1em}and\hspace{1em}}\frac{dv}{dx}={\mathrm{e}}^x$$

     then

     $$\frac{du}{dx}=2x\text{\hspace{1em}and\hspace{1em}}v={\mathrm{e}}^x$$

     Substituting these into the formula for integration by parts ...

     $$I={\mathrm{e}}^x(x^2-7)-\int 2x{\mathrm{e}}^{x}dx+C\text{\hspace{1em}[ 1 ]}$$

     The integral in this statement can be solved by repeating integration by parts

     $$u=2x\text{\hspace{1em}and\hspace{1em}}\frac{dv}{dx}={\mathrm{e}}^x$$

     then

     $$\frac{du}{dx}=2\text{\hspace{1em}and\hspace{1em}}v={\mathrm{e}}^x$$

     Substituting these into the formula for integration by parts for a second time and that into the expresssion for $I$^{[ 1 ]} then ...

     $$I={\mathrm{e}}^x(x^2-7)-[2x{\mathrm{e}}^x-2\int {\mathrm{e}}^{x}dx]+C$$ $$={\mathrm{e}}^x(x^2-7)-2x{\mathrm{e}}^x+2{\mathrm{e}}^x+C$$ $$={\mathrm{e}}^x(x^2-2x-5)+C$$

     Check by using product rule for differentiation ...

     $$\frac{dI}{dx}={\mathrm{e}}^x(x^2-2x-5)+{\mathrm{e}}^x(2x-2)$$ $$={\mathrm{e}}^x{x}^2-2x{\mathrm{e}}^x-5{\mathrm{e}}^x+2x{\mathrm{e}}^x-2{\mathrm{e}}^x$$ $$={\mathrm{e}}^x{x}^2-7{\mathrm{e}}^x$$ $$={\mathrm{e}}^x(x^2-7)$$

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2. • $$I=\int {\mathrm{e}}^x(sinx+cosx)dx$$

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     Solution to come ...

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     Answer to come ...

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Parametric Integration

Preliminary material:

Parametric Equations

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Describing Parametric Integration:

If we are given a question of the type:

$$I=\int ydx$$

and

$$x=f\left(t\right)$$ $$y=g\left(t\right)$$

where $t$ is the parameter

To answer this we can rewite the integral as

$$I=\int y\frac{dx}{dt}dt$$

Find $\frac{dx}{dt}$

Finally solving in terms of $t$

If we are carrying out definite integration then find the limits in $t$ from $t=f^{-1}\left(x\right)$

$$I={\int }_a^{b}ydx$$ $$={\int }_{t_a=f^{-1}\left(a\right)}^{t_b=f^{-1}\left(b\right)}y\frac{dx}{dt}dt$$

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Example 1:

Find

$$\int ydx$$

given

$$x=t^2+1$$ $$y=2-t^2$$

Solving as follows:

$$I=\int y\frac{dx}{dt}dt$$ $$\frac{dx}{dt}=2t$$

substituting this and $y$ into the integral

$$I=\int (2-t^2)2tdt$$ $$=\int 4t-2t^{3}dt$$ $$=2t^2-\frac{1}{2}{t}^4+C$$

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Integration Methods Flow Chart.

Remember, as with any mathematical problem, look to see if we can simplify before proceeding with any calculation or algebra. If you have an idea what the integration is, then check your possible solution by differentiation. If it's not the same as the integrand, then can it be multiplied by a constant to be correct?

If inspection doesn't work, then work through the following.

To solve a given integration problem, firstly, simplify the integrand using one or more of these:

Remove any constant factors from the integrand.

$$\int \mathrm{a}f\left(x\right)dx=\mathrm{a}\int f\left(x\right)dx$$

Split the integrand term-wise.

$$\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$$

Secondly, follow this flow chart.