Methods / ti-cas.org
	Techniques of integration + antiderivative table

To compute, the simplest case is that where the table of the antiderivatives makes it possible to directly express the primitive F of f on [ a,b ].

This article approaches some methods which go will make it possible to increase the field of the integrals which are calculable only via the table of the antiderivatives.

INTEGRATION BY PARS.

Either u and v two derivable functions on an interval I. The derivative of the product u*v is:

(u*v)’ = u’*v + u*v’,

so

u’*v = (u*v)’ – u*v’

The functions u and v are derivable, therefore continuous; If moreover functions u' and v' are continuous on I, then u'*v, (u*v) ' and u*v' are continuous thus integrable.

Either a and b two elements of I, then :

Examples

u(x)=x+1

v’(x)=e^x

u’(x)=1

v(x)=e^x

This relation is easily calculable , and

Try the prgm INTEGRPP available on the site. This program is voluntarily limited to 1 integration by parts.

INTEGRATION BY CHANGE OF VARIABLE

Change xà x+c

One proposes to calculate:

One can make calculation directly by noticing that 1/3*(x+3)^3 is an antiderivative of (t+3) ²on [ - 3,-2 ]

One can notice also that I1 is the colored surface located under the curve C1, which is equal, by translation of vector 3i, at the surface colored under C2 of equation y=x ²where 0<=x<=1

So

So

Example :

Answer :

It will be necessary to carry out an integration by parts !

Change xà c*x

One proposes to calculate:

One can directly carry out calculation by noticing that :

One can also notice that I2 is the surface colored under the curve C representative of xà e^(2x) on the interval [ 0,1 ] in the reference mark (O,i,j) where the unit of surface is the square OIKJ.

In the reference mark (O;i' ,j') where i’=1/2*i, the same curve C represents function xà e^x on the interval [ 0,2 ], "I" having for new X-coordinate 2.

So

Example :

Answer :

Change xà j (x)

One proposes to calculate:

This integral is defined on [ 1,4 ], but we do not know an antiderivative.

T appearing in in the form sqrt(t), we will take as new variable x=sqrt(t).

In the variable T, which takes values between 1 and 4, appears in and dt.

Soit j :tà sqrt(t).

j is continuous and strictly increasing on [ 1,4 ]

j (1)=1 and j (4)=2

Therefore when t varies between 1 and 4, x varies between 1 and 2.

**

The functionxà 1/(1+x) is continuous, therefore integrable on [ 1,2 ]

** x=j (t) et therefore by using the differential notation of the derivative j ’(t)=dx/dt :

dx=j ’(t)*dt, so

2*sqrt(t)*dx = dt, dt=2x*dx (because sqrt(t)=x))

One shows by using the definition of an integral and the theorem of the derivation of a made up function that :

This new integral, which does not contain any more the variable T, was obtained by changing the terminals of integration and by replacing f(t) and dt by their expressions according to X and dx.

Try the prgm CHGMTVAR available on the site.

(example with f(x)=(1-x)*sqrt(1-x²) and the change x=cos(t) )

ANTIDERIVATIVE TABLE

Olivier Miclo, october 2000.