MA 3972-MA-Book April 11, 2018 16:32AP Calculus AB Practice Exam 1 365Using area of a triangle,∫
− 1
− 3f(t)dt=−( 2 )( 2 )
2
=−2, and
∫ 0− 1f(t)dt =( 1 )( 1 )
2
=
1
2
. Therefore,
g(− 3 )=−(
− 2 +1
2
)
=3
2
.
- The correct choice is (D).
Begin with
dy
dx
= 3 x^2 yand obtain
dy
y
= 3 x^2 dx. Integrating both sides of the
equation, you have ln|y|=x^3 +c. Substitute
x=0 andy=e, and obtain ln|e|=(0)^3 +c,or
c=1, and the equation becomes ln|y|=x^3 +1.
Solving fory, you haveeln|y|=e(x^3 +1),or
|y|=e(x^3 +1), which yieldsy=±e(x^3 +1). Since
y( 0 )=e, which means the graph passes
through the point (0,e),y=e(x^3 +1).- The correct choice is (B).
Begin by solving the differential equation.
Rewrite
dy
dx=
3
x
asdy=3
x
dx. Integrating bothsides, you have∫
dy=∫
3
x
dx,or
y=3lnx+c. Substitutex=eandy=4 in the
equation, and obtain 4=3lne+c,orc=1.
Since, the point (e, 4) is on the graph off(x),
y=3lnx+1.- The correct choice is (D).
(^03)
n
y y = x
(^2) + 1
x
(^23) n 3
3 3
n
The width of each rectangle is
3 − 0
n=
3
n
, and
the heights of the rectangles are as follows:
f(
3
n)
,f(
2(
3
n))
,f(
3(
3
n))
...or
(
3
n) 2
+1,(
2(
3
n)) 2
+1,
(
3(
3
n)) 2
+1,...The area under the curve is approximately
∑nk= 13
n((
3 k
n) 2
+ 1)
, and the
∫ 30(x^2 +1)dx=lim
n→∞∑nk= 13
n((
3 k
n) 2
+ 1)
.Section I Part B- The correct answer is (D).
Since the functionf is differentiable atx=1,
f is also continuous atx=1. Atx=1, set
x^3 +x+a= 2 bx−1, and obtain
( 1 )^3 +( 1 )+a= 2 b( 1 )−1, ora= 2 b−3. Also,
f′(x)= 3 x^2 +1 forx≤1, andf′(x)= 2 bfor
x>1. Atx=1, set 3x^2 + 1 = 2 b, and you have
3 ( 1 )^2 + 1 = 2 b,orb=2. Substitutingb=2in
a= 2 b−3, you havea=1. Therefore,
a+ 2 b= 1 + 2 ( 2 )=5. - The correct answer is (D).
Note that∫ 21g′(x)dx=g( 2 )−g( 1 ).Therefore,∫ 21cos(
πx^2 +^1)
dx=g( 2 )−1
2
,org( 2 )=∫ 21cos(
πx^2 +^1)
dx+1
2
. Using your
graphing calculator, you have∫
2
1
cos(
πx^2 +^1)
dx≈ 0 .18126. Thus,g( 2 )≈ 0. 18126 +