I wonder if somebody can evaluate the following integrals
1. ∫ 1/β e^(-x/β) dx
2. ∫ x/β e^(-x/β) dx
3. ∫ e^(tx) / β e^(-x/β) dx
lets see who of you guyzz can solve it.
Suggest you to get a Derive software
i am bored with integration solving, haha
1) [e^(-x/β)]'=e^(-x/β) (-x/β)'= - 1/β e^(-x/β)
∫ 1/β e^(-x/β) dx = -e^(-x/β)+C
2) ∫ x/β e^(-x/β) dx = ∫xd(-e^(-x/β)) = -∫xd(e^(-x/β))= -xe^(-x/β)+∫e^(-x/β)dx = -xe^(-x/β) - βe^(-x/β)+ C
i am tired...
thanks for solving the problem.
Could you please tell me what techniques u used in solving those problems.
Could you please help me doing the third one..plzzzzzzzzzzzzzzzz
i would be more than gradeful
i'm not good at integration and i'm not sure whether i did it correct or not
since you wanna find its integral, or antiderivative, F(x) is one of the answers if its derivative is truely the one whose derivative you wanna find
F(x)+Cs share the same derivative with F(x), F(x)+C is also the antiderivative of F'(x)
given another one function satisfying G'(x)=F'(x) G(x) must be of the form F(x)+C
for (G(x)-F(x))'=G'(x)-F'(x)=0 and only C'=0
Conclusion: once you've found a function F(x) whose derivative is the one you wanna integrate, F(x)+c represents any its antiderivative
(Cf(x))'=C(f(x))' or=Cf'(x) where C is a constant
f'(x)=dy/dx x is any variable defined in domain
hence x itself could be a function too, as long as its value vary in the domain
it makes sense that
hence dy/dz= (f(g(z)))' of z=dy/dx dx/dz= f'(x) g'(z)=f'(g(z)) g'(z)
substitude x for z in the formula above
using tech 3 tech4 and tech5, do you understand how to solve problem 1 now?
∫ e^(tx) / β e^(-x/β) dx
do you mean it by ∫e[sup]tx/β[/sup]e[sup]-x/β[/sup] dx ?
or ∫e[sup]tx[/sup]/(β e[sup]-x/β[/sup]) dx?