I need help solving the integral of t * e^-3t dt
I know that I have to solve using integration by parts and so I've let u=t to rid of dt and dv=e^-3t which I believe would yield -(1/3) * e^-3t as v. Am I doing this correctly?
Thanks. I appreciate the help. (:
Good work. So,
∫ u dv = uv - ∫ v du
= -1/3 t e^-3t - ∫ (-1/3) e^-3t dt
and you're almost done
Awesome, thanks for the help!
Yes, you are on the right track!
To integrate the function t * e^(-3t), you can indeed use integration by parts. Let's recall the formula for integration by parts:
∫ u dv = uv - ∫ v du
In your case, you let u = t and dv = e^(-3t) dt. Differentiating u gives du = dt, and integrating dv gives v = (-1/3) e^(-3t).
Now, using the integration by parts formula, we can plug in the values:
∫ t * e^(-3t) dt = t * (-1/3) e^(-3t) - ∫ (-1/3) e^(-3t) dt
Simplifying the integral further, we have:
= -(1/3) t * e^(-3t) + (1/9) e^(-3t) + C,
where C is the constant of integration.
Therefore, the integral of t * e^(-3t) dt is equal to -(1/3) t * e^(-3t) + (1/9) e^(-3t) + C.
I hope this helps! Let me know if you have any further questions.
Yes, you are correct in using the integration by parts method to solve the given integral. Let's go through the steps together to solve it.
Integration by parts involves selecting two functions, u and dv, and finding their respective derivatives du and v to rewrite the expression as ∫ u dv = uv - ∫ v du.
In this case, you correctly selected u = t and dv = e^(-3t). Now, let's find du (the derivative of u) and v (the integral of dv).
To find du, differentiate u with respect to t:
du/dt = 1.
To find v, integrate dv with respect to t:
v = ∫e^(-3t) dt.
The integral of e^(-3t) can be found using the power rule of integration. With a constant of -3 in the exponent, we divide the integral by the constant:
v = (-1/3) ∫e^(-3t) dt.
Now, let's integrate e^(-3t) using a substitution u = -3t:
First, find the derivative du/dt:
du/dt = d(-3t)/dt = -3.
Rearranging this expression, we have dt = du / (-3).
Substituting this back into the integral, we get:
v = (-1/3) ∫e^u (du / -3).
Simplifying the integral, we now have:
v = (-1/3)(-1/3) ∫e^u du.
Integrating e^u, we find:
v = (-1/3)(-1/3) * e^u.
Plugging back in u = -3t, we have:
v = (-1/3)(-1/3) * e^(-3t).
So, after evaluating the integral, we have v = (-1/9) * e^(-3t).
Now, let's substitute these results into the integration by parts formula:
∫ t * e^(-3t) dt = (t * (-1/9) * e^(-3t)) - ∫ (-1/9) * e^(-3t) dt.
Simplifying, we get:
∫ t * e^(-3t) dt = (-1/9) * t * e^(-3t) + (1/9) * ∫ e^(-3t) dt.
Notice that the integral on the right-hand side is the same as the original integral we were trying to solve. So we have a recursive situation.
To solve for the original integral, we can rewrite it as:
∫ t * e^(-3t) dt = (-1/9) * t * e^(-3t) + (1/9) * ∫ e^(-3t) dt.
Now, let's solve for the second term on the right-hand side:
∫ e^(-3t) dt = ∫ dv.
Following the same steps as before, u = 1 and dv = e^(-3t), so du/dt = 0 and v = ∫e^(-3t) dt = (-1/9) * e^(-3t).
Substituting these values back into the equation, we get:
∫ e^(-3t) dt = (-1/9) * e^(-3t).
Now, going back to the initial equation, we can substitute this result back in:
∫ t * e^(-3t) dt = (-1/9) * t * e^(-3t) + (1/9) * (-1/9) * e^(-3t).
Simplifying further, we have:
∫ t * e^(-3t) dt = (-1/9) * t * e^(-3t) - (1/81) * e^(-3t).
Finally, we can combine the terms and simplify the result:
∫ t * e^(-3t) dt = -(1/9) * [t + (1/9)] * e^(-3t) + C,
where C is the constant of integration.
So, the solution to the integral is:
∫ t * e^(-3t) dt = -(1/9) * [t + (1/9)] * e^(-3t) + C.
I hope this explanation helps you understand how to solve the integral using integration by parts. Let me know if you have any further questions!