Question

Find the area bounded between the curve and the x-axis on the given interval.

f(x) = sec(x) on [0, π/4]

Answer 1

integrate sec(x) from 0 to pi/4

Answer 2

sec(x) was actually integrated before the integration process was even concieved

Answer 3

how can you tell if it was already integrated?

Answer 4

some map maker was wanting to do something or another, and he worked out the formula needed to integrate it

Answer 5

\[\int sec=\int sec\frac{sec+tan}{sec+tan}=\int\frac{sec^2+sec~tan}{tan~+~sec}\]

now that the top is the derivative of the bottom; this is just the derivative of a log

Answer 6

\[\int \frac{1}{u}du\]

Answer 7

when you use the interval 0, pi/4 is it the smaller value minus the big value or vice versa?

Answer 8

top minus bottom: big and small have no bearing on it
\[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]

Answer 9

the interval from a to b ...

Answer 10

i'm still not getting the answer. if it is 1/u du, then shouldn't it be ln(sec(pi/4))

Answer 11

\[ln(\frac{tan(45)+sec(45)}{tan(0)+sec(0)})\]
\[ln(\frac{1+\sqrt2}{0+1})\]
\[ln(1+\sqrt2)\]

is what i get

Answer 12

with any luck i did it right :)

Answer 13

there isnt some continuity issues in that interval is there?

Answer 14

how'd you get the ln you got: ln (tan45 + sec45/ ....)

Answer 15

well; i showed how to modify sec(x) to get a function that is of the from 1/u du

u = tan+sec

\[\int_{a}^{b}\frac1udu=ln(u(b))-ln(u(a))\]
a property of logs is that subtraction is division

\[ln(u(b))-ln(u(a))=ln(\frac{u(b)}{u(a)})\]
\[ln(\frac{tan(b)+sec(b)}{tan(a)+sec(a)})\]