For each x 0, let G(x) = ∫(from 0 to x) e^(-xt) dt. Prove...

Question

For each x > 0, let G(x) = ∫(from 0 to x) e^(-xt) dt. Prove that xG(x) = 1 for each x > 0.

Accepted Answer

Welcome back everyone. Let f of x be equal to the integral from 0 to infinity of x e to the power of negative xtdt for each x greater than 0. Find the exact value of f of x. Aus 3, B2, C1 and D 0. So let's begin this problem by rewriting the integral in terms of a limit. We can show that fx can be written as limit of the integral from 0 to b. We are replacing infinity with b, and b approaches infinity. We're going to integrate x e to the power of negative x c. The sea. So now let's begin integrating, and we can integrate this expression knowing that x is a constant first rule, and then we're going to use the u substitution. Let's suppose that u is equal to the exponent g xt. And let's remember that our next step is to find the derivative of u with respect to t. That would be the derivative of negative x multiplied by t, right? So negative x is a constant and the derivative oft is equal to 1, so we get negative x. Now if we multiply both sides by negative dt, we're going to get negative duu is equal to x dt, and this is exactly what we have within our integral x multiplied by dt. So the next step is to change the limits of integration. The lower 1 u1 is going to be neg x multiplied by 0, which is 0. And the upper limit of integration u2 is going to be neg x multiplied by b, which is basically negative bx. We can now rewrite our integral in terms of UNDu, so that'd be integral from 0 up to gbx off. E to the power of u and x multiplied by dt becomes negative du u. We can write d u and we can factor out the negative sign. From here we can apply the properties of integrals and we can swap the limits of integration. We're going to integrate from negative bx up to 0, and we're going to eliminate the negative sign. Let's now integrate e to the power of uDu. This is an elementary integral, and its value is e to the power of u. And from here we're going to apply the fundamental theorem of calculus part 2. So let's evaluate e to the power of u and 0, which is e to the power of 0, and subtract e to the power of negative bx. This is equal to 1 minus e to the power of bx can be written as 1 divided by e to the power of bx, right? So now our limit becomes limit as b approaches infinity of the result 1 minus 1 divided by e to the power of b x. We know that B, I'm sorry, we know that X is positive. So when B approaches infinity. Eats to the power of. B x is going to also approach infinity since we are going to have an infinitely large positive exponent right and e erase that exponent also gives us infinity. Now a constant divided by infinity. Gives us a limit of 0. So what is our limit? Well, that's 1 minus 0, which is 1. So this limit is one. And It basically means that the value of f x is equal to 1. The correct answer for this problem corresponds to the answer to a c. Thank you for watching.

For each x > 0, let G(x) = ∫(from 0 to x) e^(-xt) dt. Prove that xG(x) = 1 for each x > 0.

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