Consider a function f ( x ) which satisfies f ′ ( x ) = tan 2 x + ∫ − 4 π 4 π f ( x ) d x .
If f ( 4 π ) = − 4 π , evaluate ⌈ − 1 0 0 0 ∫ − 4 π 4 π f ( x ) 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.
Problem Loading...
Note Loading...
Set Loading...
Let C = ∫ − 4 π 4 π f ( x ) d x , which is a constant. Then
f ′ ( x ) = tan 2 ( x ) + C = sec 2 ( x ) − 1 + C ⟹ f ( x ) = tan ( x ) − x + C x + D
for some constant D . We are then given that
f ( 4 π ) = tan ( 4 π ) − 4 π + 4 π C + D = − 4 π ⟹ 4 π C + D = − 1 , (i).
Now when we integrate f ( x ) = tan ( x ) − x + C x + D from − 4 π to 4 π , since tan ( x ) and x are odd functions they will contribute nothing to this symmetric integral, so we are left to evaluate D x from the lower to upper bound, giving us that C = 2 π D . Plugging this result into equation (i) yields that
4 π × 2 π D + D = − 1 ⟹ D ( 1 + 8 π 2 ) = − 1 ⟹ D = − 8 + π 2 8 ⟹
C = 2 π D = − 8 + π 2 4 π = − 0 . 7 0 3 2 2 6 to 6 decimal places.
Thus ⌈ − 1 0 0 0 × C ⌉ = ⌈ 7 0 3 . 2 2 6 ⌉ = 7 0 4 .