Integral and Derivative

Calculus Level 4

Consider a function f ( x ) f(x) which satisfies f ( x ) = tan 2 x + π 4 π 4 f ( x ) d x \displaystyle f'(x) = \tan^2{x} + \int_{-\frac{\pi}{4}}^\frac{\pi}{4} f(x) \, dx .

If f ( π 4 ) = π 4 f \left(\dfrac{\pi}{4}\right) = -\dfrac{\pi}{4} , evaluate 1000 π 4 π 4 f ( x ) d x \displaystyle \left \lceil -1000\int_{-\frac{\pi}{4}}^\frac{\pi}{4} f(x) \, dx \right \rceil .


The answer is 704.

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1 solution

Let C = π 4 π 4 f ( x ) d x C = \displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) dx , which is a constant. Then

f ( x ) = tan 2 ( x ) + C = sec 2 ( x ) 1 + C f ( x ) = tan ( x ) x + C x + D f'(x) = \tan^{2}(x) + C = \sec^{2}(x) - 1 + C \Longrightarrow f(x) = \tan(x) - x + Cx + D

for some constant D D . We are then given that

f ( π 4 ) = tan ( π 4 ) π 4 + π 4 C + D = π 4 π 4 C + D = 1 f\left(\dfrac{\pi}{4}\right) = \tan\left(\dfrac{\pi}{4}\right) - \dfrac{\pi}{4} + \dfrac{\pi}{4}C + D = -\dfrac{\pi}{4} \Longrightarrow \dfrac{\pi}{4}C + D = -1 , (i).

Now when we integrate f ( x ) = tan ( x ) x + C x + D f(x) = \tan(x) - x + Cx + D from π 4 -\dfrac{\pi}{4} to π 4 \dfrac{\pi}{4} , since tan ( x ) \tan(x) and x x are odd functions they will contribute nothing to this symmetric integral, so we are left to evaluate D x Dx from the lower to upper bound, giving us that C = π 2 D C = \dfrac{\pi}{2}D . Plugging this result into equation (i) yields that

π 4 × π 2 D + D = 1 D ( 1 + π 2 8 ) = 1 D = 8 8 + π 2 \dfrac{\pi}{4} \times \dfrac{\pi}{2}D + D = -1 \Longrightarrow D\left(1 + \dfrac{\pi^{2}}{8}\right) = -1 \Longrightarrow D = -\dfrac{8}{8 + \pi^{2}} \Longrightarrow

C = π 2 D = 4 π 8 + π 2 = 0.703226 C = \dfrac{\pi}{2}D = -\dfrac{4\pi}{8 + \pi^{2}} = -0.703226 to 6 decimal places.

Thus 1000 × C = 703.226 = 704 \lceil -1000 \times C \rceil = \lceil 703.226 \rceil = \boxed{704} .

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