Question

99 POINT QUESTION, PLUS BRAINLIEST!!!
(Please answer genuinely, and do not answer just for points, if you do, your answer will be deleted, and those points will be taken back...)
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THIS IS CALCULUS NOT BASIC MATH...

4.) Find the area of the region bounded by the graphs of the equations.
f(x) = (16x)/((x^2)+1), y=0, 0 ≤ x ≤ 8
A • A = 8 ln (65)
B • A = 8 ln (63)
C • A = 65 ln (8)
D • A = 63 ln (8)

6.) Find the area of the region bounded by the graphs of the algebraic functions.
f (x) = x^2 + 12x + 36
g(x) = 8(x + 6)
A • A = 364/3
B • A = 128/3
C • A = 256/3
D • A = 512/3

** Please show all of you work, if you do not, your answer will be deleted, and the points you earned will be taken away...
** If you can only answer one of the questions, that's okay, just please show me how you solved it...

Answer 1

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

Answer 2

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

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