subject: Integrating Exponential Functions [print this page] The beginnings of 'Integral calculus' can be traced back to antiquity. Ancient mathematicians of Greece developed the method of exhaustion which they had applied to calculate areas of plane surfaces and volumes of solids.
Wallis ( 1616 - 1703 ) chief contribution to the development of calculus in its early period lay in the theory of integration.
Bernhard Riemann ( 1826 - 1866 ) began with the question ' when is a function integrable? ' which led him to the investigation of convergence of Cauchy sums. Thus he refined and clarified the notation of integral and that is what we call now the "Riemann Integral".
This performance is utilized to execute the combination found on the result. If individual of the yield is agreement followed by the combination on the invention is able to be simply integration. If the invention of integration is of two dissimilar types of purpose then we use the conception of integration by parts. The subsequent are the technique of integration by parts.
Trigonometric functions are represented as circular functions of an angle. They are used to relate the position of a triangle towards the length of the surface of a triangle. The sine function obtains an angle and length of the y-component of that triangle. The cosine function obtains an angle and length of x-component of a triangle. The tangent function obtains an angle and slope of y-component divided by the x-component. Now we study about trigonometric integration.
Integration of exponential Functions:-
We know that 'int' e^x dx = e^x + c , c is a constant
Let us evaluate 'int' xe^x dx on R
we take u( x ) = x and v( x ) = e^x
Now, by the formula for integration by parts, we have
'int' xe^x dx = u( x ) v' ( x ) - 'int' u' ( x ) v ( x ) dx
= x e^x - 'int' e^x dx
= x e^x - e^x + c
= ( x - 1 ) e^x + c
Let us show that, given a differentiable function f on I,
'int' e^x [ f ( x ) + f ' ( x ) ] dx = e^x f ( x ) + c
For this purpose, since [ e^x f ( x ) ] ' = e^x f ' ( x ) + e^x f ( x )
= e^x [ f ( x ) + f ' ( x ) ]
we have by the definition of the indefinite integral, it follows that
'int' e^x [ f ( x ) + f ' ( x ) ] dx = e^x f ( x ) + c
And example of Integrating e^xponential Functions :-
Let us find 'int' [ e^x ( 1 + x ) / ( 2 + x )^2 ] dx on I 'sub' R { -2 }
Solution:- We have ( 1 + x ) / ( 2 + x )^2 = ( ( 2 + x ) -1 ) / ( 2+ x )^2
= ( 2 + x ) / ( 2 + x )^2 - 1 / ( 2 + x )^2
= 1 / ( 2 + x ) - 1 / ( 2 + x )^2
Define f ( x ) = 1 / ( 2 + x ) so that f ' ( x ) = - 1 / ( 2 + x )^2
By the above formula
e^x [ ( 1 + x ) / ( 2 + x )^2 ] dx = e^x [ f ( x ) + f ' ( x ) ] dx
