12. b 3
2 M(b a)
______
2
1 ∫ f(x)dx Trap ≤
∫ 2 dx n
12n 2
1 x a
b 3
M(b a)
______
2
∫ f(x)dx Trap
Mid n ≤ 2
12n
24
a
(b) midpoint sum? 71
Only thing that = 35.5
changed is the divisor.
So, instead of n = 71,
2 (round up)
divide by 2 and round
up (since n must be a
positive integer)
n = 36
14. Find these derivatives: d
dx
sin 1
( x)
1
1
2
1 2 x Applying chain rule, we get
√1 ( ) x
our final answer to be
1
1 2
x 2
√1 ( ) x
15. Find these derivatives: d 1 arctan x
dx 4
( 4)
1 x
4 [ d arctan
dx
( 4) ] Again, applying chain rule,
we get our answer to be:
16 (1
1
+
1
( x 2
4) )
16. d
Find these derivatives: arc cot(x)
dx
Let First we make a
arc cot(x) =y substitution
Then take the cot of both
cot(arccot(x)) = cot(y) sides to solve for x
Then we differentiate
x = cot(y) both sides
1 = csc (y)2
y '
(Continues next page...)
17. 1 = 2
csc (y) y '
1 Rearrange equation to solve for y`
y = '
2
csc (y) Then input what csc (y) is 2
1 √1 + x 2
'
y = 1
2
1
( √1 + x ) 2 y
x
y = '
1x 2
19. Find these antiderivatives:
dx
∫ 4+x 2
First, we factor
1
= ∫4
1
1+ 4
x(2
)
dx out 1/4 then 2
make x /4 to be
2
(x/2) so we can
1 1
( )
= 4 ∫ 1 + ( x ) 2 dx antidifferentiate
2
the quot;thingquot; into
1
1 1 ( x ) tan and place
= 4 tan 2 + c (x/2) back in
20. dx
∫ √4 x 2
dx First factor out 4
√
=∫ x2
2√
1 √4 take the square root
of x 2/2 and square it,
1 1 to make it
= 2 ∫ 1 x 2 dx differentiateable
√ (√2 ) antidifferentiate the
1 arc sin x
(√2 ) + C quot;thingquot;
= 2
21. Find these antiderivatives:
dx
∫ x 2 + 4x + 5
quot;Complete the
1 squarequot; or add 0
=∫ dx to make a
x2 +4 x + 5 1 + 1 difference of
1 squares
= ∫ (x + 2)
2
+ 1
dx
Antidifferentiate
into arctan and
= arctan(x+2) + C substitute x+2
back in