# Evaluate the limit of I(a), if a goes to infinite. I(a) is definite integral of (2x-3)*e^x, ifx=-a to x=0

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First I(a) needs to be determined. That is given as the definite integral of (2x - 3)*e^x for x = -a to x= 0

Int [ (2x - 3)*e^x dx]

use integration by parts.

Int[ u dv ] = u*v - Int [ v du]

2x - 3 = u , du = 2*dx

dv = e^x dx, v = e^x

Int [ (2x - 3)*e^x dx]

=> (2x - 3)*e^x - Int [ 2*e^x dx]

=> (2x - 3)*e^x - 2*e^x + C

=> (2x - 5)*e^x + C

Between the limits x = -a and x = 0

(2*0 - 5)*e^0 + C - (2*(-a) - 5)*e^(-a) - C

=> -5 - (2*(-a) - 5)*e^(-a)

=> -5 - (2/a*e^a) + 5/e^a

For a--> inf , 1/e^a --> 0

-5 - (2/a*e^a) + 5/e^a

=> - 5 - 0 + 0

=> -5

**The required limit of I(a) is -5**

To evaluate the limit, we'll have to determine the definite integral first.

We'll use Leibniz-Newton formula for finding the integral:

Int (2x-3)*e^x dx = F(0)-F(-a)

We'll integrate by parts.

Int udv = uv - Int vdu

Let u = 2x-3 => du=2dx

Let dv = e^x dx => v = e^x

Int (2x-3)*e^x dx = (2x-3)*e^x - 2Int e^xdx

Int (2x-3)*e^x dx = (2x-3)*e^x - 2 e^x

Int (2x-3)*e^x dx = (e^x)*(2x - 3 - 2)

Int (2x-3)*e^x dx = (e^x)*(2x - 5)

F(0) = (e^0)*(2*0 - 5) = -5

F(-a) = (e^-a)*(-2a - 5)

F(0)-F(-a) = -5 - (e^-a)*(-2a - 5)

I(a) = F(0)-F(-a) = (e^-a)*(2a + 5) - 5

Now, we'll evaluate the requested limit, if a approaches to infinite:

lim I(a) = lim [(e^-a)*(2a + 5) - 5]

We notice that we'll get an indetermination, "infinite/infinite", therefore we'll use L'Hospital's rule:

lim [(e^-a)*(2a + 5) - 5] = lim (2a+5)'/(e^a)' - lim 5

lim (2a+5)'/(e^a)' - lim 5 = lim 2/e^a - lim 5 = 2/infinite - 5

lim I(a) = 0 - 5

lim I(a) = -5

**The requested limit of I(a), if a approaches to infinite, is: lim I(a) = -5.**