Suppose that a continuous function f ( x ) satisfies the relation ∫ x x + 1 f ( t ) d t = e x for every x ≥ 0 . Find f ( 2 ) − f ( 0 ) .
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.
Dang it, I've never seen this type of problem before. Thanks!
Let f ( t ) = A e t so that the above definite integral computes to:
∫ x x + 1 A e t d t = A e t ∣ x x + 1 = A e x + 1 − A e x = A e x ( e − 1 ) = e x
or A = e − 1 1 ⇒ f ( x ) = e − 1 e x . Thus, f ( 2 ) − f ( 0 ) = e − 1 e 2 − 1 = e − 1 ( e + 1 ) ( e − 1 ) = e + 1 .
The given integral from x to x+1 plus the same integral from x+1 to x+2 equals exp(x) + exp(x+1).
f(2)-f(0) = exp(0) + exp(1) = 1+e, because we set x = 0.
Problem Loading...
Note Loading...
Set Loading...
∫ x x + 1 f ( t ) d t ∫ 0 x + 1 f ( t ) d t − ∫ 0 x f ( t ) d t f ( x + 1 ) − f ( x ) f ( 2 ) − f ( 1 ) f ( 1 ) − f ( 0 ) ⟹ f ( 2 ) − f ( 0 ) = e x = e x = e x = e . . . ( 1 ) = 1 . . . ( 2 ) = e + 1 Differentiate both sides w.r.t. x Putting x = 1 Putting x = 1 = 0 Adding the two equations ( 1 ) + ( 2 )