Friday, January 11
Clicker Questions
Clicker Question 1
A definite integral
Z
e2
1
dt.
t
Z
e2
Evaluate
1
A. 1 −
1
e4
B. 2, since
1
e 2
1
dt = ln |x| = ln(e2 ) − ln(1) = 2 − 0
t
1
2
C. ln(e ) − 1
1
−1
e2
E. none of the above
D.
Clicker Question 2
Checking up on FTC part 1
Z
If h(x) =
x5
t3 dt, calculate h0 (x).
7
A. x3
B. 5x4
C.
x15
D.
5x19
E. 3x2
The Chain Rule again
Z
Write g(x) =
x
t3 dt, so that g0 (x) = x3 by FTC
7
part 1. Since h(x) = g(x5 ), we see that
h0 (x) = g0 (x5 ) · 5x4 = (x5 )3 · 5x4 = 5x19 .
Clicker Question 3
An indefinite integral
Z
If a > 0 is a constant, what is
A.
ax dx?
ax+1
+C
x+1
B. ax + C
ax+1
+C
a+1
1 x
d
1 x
1 d(ax )
1
D.
a + C, as
a +C =
=
(ln a)ax
ln a
dx ln a
ln a dx
ln a
C.
E. (ln a)ax + C
Clicker Question 4
What’s the right question?
√
Which of the following indefinite integrals equals x 2x + 3 + C ?
Z
A.
Z
B.
Z
C.
Z
x
dx
2x + 3
√ √
3 x + 3 dx
√
√
2x + 3 dx
3x + 3
√
dx
2x + 3
E. none of the above
D.
Indefinite integrals: sometimes
easier to check than find!
d √
x 2x + 3 + C
dx
√
1
2
= 1 · 2x + 3 + x · √
2 2x + 3
1
2
2x + 3
+x· √
=√
2
2x + 3
2x + 3
3x + 3
=√
2x + 3