For Exercises 127 and 128 find a function f satisfying each eq...

Question

For Exercises 127 and 128 find a function f satisfying each equation.

128. f(x) = e² + ∫₁ˣ f(t) dt

Accepted Answer

Hello there. Today, we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. Find a function F satisfying F of X is equal to or multiplied by E to the power of 3, plus the integral evaluated from 2 to X of F of TDT. OK. So it appears for this particular problem, we're asked to find a function F that satisfies the given function for F of X. So, now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to differentiate both sides of F of X. So we need to go ahead and write that F of X is equal to 4 multiplied by E to the power of 3. Plus the integral evaluated from 2 to X of F of T DT fantastic with respect to X. So with respect to X, which I'll denote as WRT with respect to X. So once again, we're differentiating both sides of our function for F of X with respect to X. And we need to note, based on our prior knowledge, the left side will give us F of X, and by the fundamental theorem of calculus, as we should recall, we'll be able to write that D by DX of or multiplied by E to the power of 3. Will be equal to 0 and D by DX of the integral evaluate from 2 to X of F T DT will be equal to F of X. So therefore, we can safely state that F of X is equal to F of X. Fantastic. So, our second step now is we need to rewrite F of X is equal to F of X. So once again we need to rewrite F of X is equal to F of X as DF so once again we're gonna need to take DF by DX is equal to F of X. Fantastic. So once again, We took F of X is equal to F of X, as DF by DX is equal to F of X, and we now need to separate the variables, meaning, as we should recall, we need to pull out all the function F terms on one side and all the X terms on the other. So we need to go ahead and write 1 divided by F of X. DF is equal to DX and we now need to integrate both sides. When you take the interval of 1 divided by F of X, DF is equal to the integral of DX. And we need to recall that the integral on the left will be the natural log of the absolute value of F of X. Fantastic, and we need to note that anything inside of absolute value symbols or absolute value signs will become a positive value. So once again, the integral on the left will be the natural log of the absolute value of F of X. And the integral on the right will be X plus C1 or C subscript 1, where C1 in this particular case is an integration constant value. So once again, C1, C subscript 1, C1 for short, is an integration constant. So you can go ahead and write that the natural log of the absolute value of F of X. Is equal to x + C1. So with that said, in order to solve for F of X, we need to exponentiate both sides in order to get rid of the natural logarithms, the natural logs. So you need to take the absolute value of F of X is equal to E to the power of X plus C1, which is equal to E power of C1 multiplied by E power of X. Fantastic. So, moving right along here, we need to note that since E to the power of C1 is constant, we could just simply denote it as C. So we also need to recall a note, based on our prior knowledge that the absolute value can be removed if we allow C to be any real constant, whether or not it be a positive or negative value. So we can go ahead and write that F of X. So once again we can write that F of X is equal to C multiplied by E to the power of X. Fantastic. Our 3rd step now is we need to determine the value of C by recalling and using the original equation at X equals 2, so meaning we need to plug in X equals 2 into our given equation. So meaning we need to take F of 2, plugging in a value of 2 in for X. So F of 2 is equal to 4 multiplied bye the power of 3 plus 0, which is equal to 4 multiplied by E to the power of 3. So from the general form that we saw in step 2, we can go ahead and write that F of 2 is equal to C multiplied by E to the power of 2. So now we need to equate these expressions, so we need to take C multiplied by E 2 is equal to 4 multiplied by E3, and rearranging this expression to isolate and solve for C, we will find that C is simply equal to 4 multiplied by E 3 divided by E2, which simplifies to equal 4 multiplied by E. So Therefore, without a doubt, we can state that F of X will be equal to 4 multiplied by E, multiplied by E to the power of X, which will be equal to 4 multiplied bye the power of X + 1 when we simplify our expression, and this will be our final answer. Hooray, we did it. So looking back at the problem itself to recap what we've learned. We can safely say that our final answer once again without a doubt will be. F of X is equal to or multiplied by e to the power of X + 1, and that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

For Exercises 127 and 128 find a function f satisfying each equation.128. f(x) = e² + ∫₁ˣ f(t) dt

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For Exercises 127 and 128 find a function f satisfying each equation.
128. f(x) = e² + ∫₁ˣ f(t) dt