Hi I'm having a bit of trouble if anyone is able to help, thanks!
Using the substitution u =1-2t^2, show that
A = -1/4 ∫_1^(-7)▒e^u du
To solve this problem, we need to use the given substitution u = 1 - 2t^2.
First, let's find the derivative du/dt by differentiating both sides of the equation u = 1 - 2t^2 with respect to t:
du/dt = d(1 - 2t^2)/dt
du/dt = -4t
Now, we can solve for dt by rearranging the equation:
dt = du / (-4t)
Next, we need to find the new limits of integration. When t = 1, we substitute into the equation u = 1 - 2t^2:
u = 1 - 2(1^2)
u = 1 - 2
u = -1
When t = -7, we substitute into the equation u = 1 - 2t^2:
u = 1 - 2(-7^2)
u = 1 - 2(49)
u = 1 - 98
u = -97
Now, let's substitute the values for du/dt, dt, and the new limits of integration into the original integral:
A = -1/4 ∫_1^(-7) e^u du
Using the substitution properties, we can rewrite this integral as:
A = -1/4 ∫_(-1)^(-97) e^u du
Now, we can integrate with respect to u:
A = -1/4 [e^u]_(-1)^(-97)
A = -1/4 (e^(-97) - e^(-1))
So, the final answer is:
A = -1/4 (e^(-97) - e^(-1))
I hope this explanation helps! Let me know if you have any further questions.