1 0 x + 1 1 x + 1 2 x = 1 3 x + 1 4 x
If A is the number of solutions of the equation above, find ∫ 0 1 x A e d x .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
I know that A is 1, but how do we get to the answer of the real question?
Log in to reply
just integrate
e + 1 1 is the answer of the integrand.
I don't remember how to, and it can't be a solution if it's not showing the final answer in it.
Problem Loading...
Note Loading...
Set Loading...
1 3 x 1 0 x + 1 3 x 1 1 x + 1 3 x 1 2 x = 1 + 1 3 x 1 4 x On Treating the LHS and RHS separately we find that the LHS is a decreasing function and the RHS is an increasing function therefore their graphs can have at most 1 intersection point. which through inspection lies at x = 2 . Note for the problem you weren't required to find the value which satisfies the function.