what is the answer

Question

Integrate ∫₀ ⁷ tan x dy

A. tan 7 - tan 0

B. ln | cos0 | -ln | cos7

C. 7 tan x

D. Cannot integrate because the function is not continous from 0 to 7.

Answer 1

Integral from 0 to 7 of tanx dx (not dy)

Assume radians.

The graph of tanx from 0 to 2pi (that's ~ 6.28 radians)

Shows an equal area above and below the x-axis

So, the positive and negative areas cancel.

So, the integral from 0 to 2pi of tanx is zero.

All we have to do is to find the integral from 2pi to 7

And notice that part is all under the x-axis

So we should expect a negative value.

Original Integral

= Integral from x=2pi to x=7 of tanx dx

= Integral from x=2pi to x=7 of 1/cosx * sinx dx

= - Integral from x=2pi to x=7 of 1/u * du With u = cosx, du = -sinx dx

= - [ ln| u | ]

= - [ ln| cosx| ] from x=2pi to x= 7

= - [ ln| cos(7) | - ln| cos(2pi) | ]

= - [ ln( cos(7) ) - 0 ]

= - ln( cos(7) )

Absolute Value sign not needed because cos7 = positive

Also, ln( cos(2pi) ) = ln( 1 ) = 0

So, the answer is B.

(But the answer could be simplified.)

Answer 2

Answer: D (Assuming that "dy" is supposed to be "dx")

tanx is undefined at π/2 ≈ 1.57 and 3π/2 ≈ 4.71, for example.

Note: if "dy" is not a typo, then C would be the correct answer, since tanx would be treated as constant and would be multiplied by the integral of dy from y = 0 to y = 7.