Question

99 POINT QUESTION, PLUS BRAINLIEST!!!

Answer 1

The first solution:
$\int\limits^8_0 {\frac{16x}{1+x^2}} \, dx=8log(65)$

The second solution:
$\int\limits^2_{-6}{(-36-12x-x^2+8(6+x))} \, dx=\frac{256}{3}$

Answer 2

4.
first one, since the lower bound is y=0

so what we do is just inetgrate from x=0 to x=8


$\int\limits^8_9 { \frac{16x}{x^2+1} } \, dx$
find an antiederivitive

use u subsitution
u=x²+1
du=2x dx
so factor out the 8

$8 \int\limits^8_9 { \frac{2x}{x^2+1} } \, dx$
$8 \int\limits^8_9 { \frac{1}{u} } \, du$
and we know that the antideritivitve of 1/u is ln|u|
8ln|x²+1| is an antideritivive
now
$[8ln|x^2+1|]^8_0$
$8ln|8^2+1|-8ln|0^2+1|$
$8ln|65|+8ln|1|$
$8ln65+0$
the aera is 8ln65
A is the answer





6.
find where they intersect to find the area bounded
f(x)=g(x) at x=-6 and x=2
and g(x) is on top (we can see that because g(0)>f(0))
so we integrate from -6 to 2
$\int\limits^2_{-6} {g(x)-f(x)} \, dx$
$\int\limits^2_{-6} {8x+48-(x^2+12x+36)} \, dx$
$\int\limits^2_{-6} {8x+48-x^2-12x-36)} \, dx$
$\int\limits^2_{-6} {-x^2-4x+12)} \, dx$
this is easy
use reverse power rules
remember that $\int\limits^a_b {f(x)+g(x)} \, dx = \int\limits^a_b {f(x)} \, dx + \int\limits^a_b {g(x)} \, dx$
the antiderivitive is
$\frac{-x^3}{3} -2x^2+12x$
so
 $[\frac{-x^3}{3} -2x^2+12x]^2_{-6}$=
$(\frac{-2^3}{3} -2(2)^2+12(2))-(\frac{-(-6)^3}{3} -2(-6)^2+12(-6))$=
$({-8}{3} -8+24)-(72 -72-72)$=
$({-8}{3}+ 16)-(-72)$=
${-8}{3}+ 16+72$=
${-8}{3}+ 88$=
$\frac{-8}{3} + \frac{264}{3}$=
$\frac{256}{3}$

answer is C