sqrt2 times e^t times t^(1/2) dt bounded between 0 and 1.
i tried integration by parts.. but it keeps repeating.. please help. thank you!
assistance needed
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Let's leave out the constant sqrt2 and do the rest. The constant can be multiplied back later.
I will use S to represent the integral sign.
Let u = e^t ; du = e^t dt
Let v = t^(1/2) ; dv = (1/2) t^(-1/2)
S u dv = e^t*t^(1/2) - S t^(1/2) e^t dt
= e^t*t^(1/2) - S u dv
Therefore
2 S u dv = e^t*t^(1/2)
S u dv = (1/2) e^t*t^(1/2)
Multiply that by sqrt2 for the integral you were asked for.
To solve the integral of √2 * e^t * t^(1/2) dt bounded between 0 and 1, it is possible to apply integration by parts multiple times. However, in this case, it would be more efficient to use the substitution method.
Let's start by applying the substitution u = t^(1/2). To find du, differentiate both sides of the equation with respect to t:
du/dt = (1/2) * t^(-1/2)
Now, we can solve for dt:
dt = 2u * du
Substitute the expression for dt back into the integral:
∫[0 to 1] √2 * e^t * t^(1/2) dt = ∫[0 to 1] √2 * e^t * t^(1/2) * 2u du
Simplifying the integral:
2√2 * ∫[0 to 1] e^t * u du
Now, let's focus on the integral: ∫[0 to 1] e^t * u du. We can now integrate this by parts:
Let dv = e^t dt
Then, v = e^t (using the integral property of e^t)
Let u = u
Then, du = du
Using the integration by parts formula:
∫ udv = uv - ∫ vdu
Apply these substitutions:
∫[0 to 1] e^t * u du = [e^t * u] from 0 to 1 - ∫[0 to 1] (e^t) * du
Simplifying:
= e^1 * 1 - e^0 * 0 - ∫[0 to 1] e^t dt
= e - ∫[0 to 1] e^t dt
Now, we just need to solve the integral of e^t from 0 to 1, which is a standard integral:
∫[0 to 1] e^t dt = [e^t] from 0 to 1
= e^1 - e^0
= e - 1
Substituting this result back into the previous expression:
= e - (e - 1)
= 1
So, the final answer to the integral ∫[0 to 1] √2 * e^t * t^(1/2) dt is 2√2.