Find the exact value of integral t^2(2)^(-t^3). Upper limit is 1 and lower limit is 0.
if you let u=-t^3 then
du = -3t^2 dt
So now you have ∫-1/3 2^u du
Now recall that ∫a^u du = 1/lna a^u + C
To find the exact value of the integral ∫[0,1] t^2 * 2^(-t^3), we can use the technique of integration by substitution.
Let's make the substitution u = -t^3. To find du, we differentiate both sides of the equation with respect to t:
du = -(3t^2) dt
Now, we can rewrite the integral in terms of u:
∫[0,1] t^2 * 2^(-t^3) dt
= ∫[0,1] t^2 * 2^u dt
Next, we need to substitute dt in terms of du. From the equation du = -(3t^2) dt, we can solve for dt:
dt = -(1/3t^2) du
Now our integral becomes:
∫[0,1] t^2 * 2^u * (-(1/3t^2)) du
Simplifying further:
∫[0,1] (1/3) * 2^u du
Now, let's solve this integral. We can go back to the original variable t by solving t from the substitution u = -t^3:
u = -t^3
-t^3 = u
t = (-u)^(1/3)
We also need to consider the limits of integration when we substitute back:
When t = 0, u = -0^3 = 0
When t = 1, u = -1^3 = -1
Therefore, the new integral expression becomes:
∫[-1,0] (1/3) * 2^u du
Using the formula for integrating an exponential function, we have:
∫[-1,0] (1/3) * 2^u du
= (1/3) * [2^u / ln(2)] | from -1 to 0
= (1/3) * (2^0 / ln(2) - 2^(-1) / ln(2))
= (1/3) * (1 / ln(2) - 1/2 / ln(2))
= (1/3) * (1/ln(2) - 1/(2ln(2)))
= (1/3) * (2 - 1/(2ln(2))) / ln(2)
Thus, the exact value of the integral ∫[0,1] t^2 * 2^(-t^3) is (1/3) * (2 - 1/(2ln(2))) / ln(2).