What is the smallest value of a if the area under the curve y=e^(ax) from -infinity to 4 is 11?

Question

What is the smallest value of a if the area under the curve y=e^(ax) from -infinity to 4 is 11? I cannot seem to arrive at the answer using integrals. Please help.

Answer 1

The solution of Ketaki is correct. There are two solutions to the equation e^4a = 11a. That is why there is the smallest value of a. Pick the smaller a in the solution.

 

The curve 11a starts at 0 when a = 0 while the curve e^4a = 1 when a = 0. So the curve y = 11a is below the curve

y = e^4a. For a = infinity the curve y = 11a is also below the curve y = e^4a. So if the two curves intersect they have to cross at two points (unless the line is tangent to the curve, but not in this case).

 

If you graph the two curves carefully, you will see that they cross at two points.

Answer 2

Indefinite integral of e^ax is e^ax/a ..

 

when we apply the limits from -inf to 4 we get( e^4a - e^-inf)/a

 

now i am assuming a to be positive ,otherwise the area would be infinite.

 

 so e^-inf tends to 0. 

 

the required area is e^4a/a= 11 

 

now I think you will have to use scientific calculator to find the value of a that satisfies this equation