# Thread: exponential functipn definite integral

1. ## exponential function definite integral

I have a function:

f(x) = 10.767(1.346)^x

I have found the area below the graph between x = -8, and x= 0, below y = 8. This comes out to 31.557.

I need to change my function f(x) so that the area between x -8, x = 0, below y = 8 is equal to different areas.

The areas I am testing are:

1.25 (31.557)
1.5(31.557)
2(31.557)

How would I go about doing this?, It's been bugging me.

2. What exactly are you changing in f(x)?

3. a and b

4. $f(x) = 10.767(1.346)^x$

I don't see an a or b in that equation. Are you talking about the intervale from -8 to 0?

5. Originally Posted by Mr.Ree
I have a function:

f(x) = 10.767(1.346)^x

I have found the area below the graph between x = -8, and x= 0, below y = 8. This comes out to 31.557.

I need to change my function f(x) so that the area between x -8, x = 0, below y = 8 is equal to different areas.

The areas I am testing are:

1.25 (31.557)
1.5(31.557)
2(31.557)

How would I go about doing this?, It's been bugging me.
Do you need this

$10.767{\left( {1.346} \right)^x} = 8 \Leftrightarrow {\left( {1.346}\right)^x} = \frac{8}{{10.767}} \Leftrightarrow x = {\log _{1.346}}\frac{8}{{10.767}} \approx - 1.$

$A = \int\limits_{ - 8}^{ - 1} {10.767{{\left( {1.346} \right)}^x}dx} + \int\limits_{ - 1}^0 {8dx} = \left. {\frac{{10.767}}{{\ln 1.346}}{{\left( {1.346} \right)}^x}} \right|_{ - 8}^{ - 1} + \left. {8x} \right|_{ - 1}^0 \approx 31.55769686$