Question

Number 14 on section 5.5 of the book. " calculus. Early transcendentals tenth edition"

Answer 1

Can you copy the question to here?

Answer 2

@RyanRyan

Answer 3

Is it an integration problem?

Answer 4

It's an integration problem.
" sketch the region whose signed area is represented by the definite integral, and evaluate the integral using the approximate formula from geometry, where needed "..
Integral sign with a 2 on top and 0 on bottom ( 1-1/2x)dx

Answer 5

Yes

Answer 6

\[\int\limits_{0}^{2} (1-\frac{1}{2}x)dx = \frac{1}{2} \times 2 \times 1 = 1\]

Answer 7

yes abb0t, that is it. I am having to do a,b,c, and d.

Answer 8

trapezoid is the same thing, except evaluated from -1<x<1 with base 2 and parrallel sides 1.5 and .5 so it's \[\int\limits_{-1}^{1}(1-\frac{1}{2}x)dx\]

Answer 9

triangle is \[\int\limits_{2}^{3}(1-\frac{1}{2})dx = 1 \times 0.5\] and last one is \[\int\limits_{0}^{3}(1-\frac{1}{2})dx \] which you can do yourself.

Answer 10

thanks, how did you get the answers so quick? Are you using the Riemann Summ??

Answer 11

No. \[\int dx - \int (\frac{1}{2}x)dx\] i separated them
\[\int_{a}^{b} dx = x,~~evaluated~ ~from~~a<x<b\] and you know that by the fundamental theorem of calculus, that you simply use the difference.
then for the second integral \[\int_{a}^{b} \frac{1}{2}xdx = \frac{1}{4}x^2\] also evaluated from a<x<b

where a and b are your bounds.

Answer 12

thanks