FIVE MINUTE REVIEW FOR WEEK 2.
Question 1. Factor 4x2 − 9.
Answer 1.
4x2 − 9 = (2x + 3)(2x − 3)
Question 2. Simplify 4(x2 − x + 2) − 5(x − 1)2 .
Answer 2.
4(x2 − x + 2) − 5(x − 1)2 = 4x2 − 4x + 8 − 5(x2 − 2x + 1) = 4x2 − 4x + 8 − 5x2 + 10x − 5 = −x2 + 6x + 3
Question 3. Simplify
1
x+1
1
x−1 .
+
Answer 3.
1
x−1+x+1
2x
1
+
=
= 2
x+1 x−1
x2 − 1
x −1
3
Question 4. Simplify (x−5 y 3 z 10 )− 5 .
Answer 4.
3
x3
9
(x−5 y 3 z 10 )− 5 = x3 y 5 z −6 =
9
y 5 z6
Question 5. Solve 10θ2 − eθ + 10 = 0.
Answer 5.
θ=
Question 6. Calculate
d
dx ((x
e±
√
d
dθ (log(cos θ)).
d
− sin θ
(log(cos θ)) =
= − tan θ
dθ
cos θ
d
dt
Rt
√
2
2
esin(s ) ds.
Answer 8.
d
dt
Question 9. Calculate
Z
t
√
2
2
esin(s ) ds = esin(t
)
2
d
2
dx (arcsin(x )).
Answer 9.
Question 10. Calculate
Answer 10.
(e − 20)(e + 20)
20
d
((x − 2)3 ) = 3(x − 2)2
dx
Answer 7.
Question 8. Calculate
p
− 2)3 ).
Answer 6.
Question 7. Calculate
e±
e2 − 400
=
20
d
2x
(arcsin(x2 )) = √
dx
1 − x4
d
2
dx (x
log(x)).
d 2
(x log(x)) = 2x log(x) + x
dx
Date: August 31, 2015.
1
2
FIVE MINUTE REVIEW FOR WEEK 2.
Question 11. Calculate
Rφ
dθ
.
0 cos2 θ
Answer 11.
φ
Z
0
Question 12. Calculate
Answer 12.
R
π
2
Z
π
2
π
4
sin θ
cos θe
dθ.
cos θesin θ dθ =
π
4
dθ
= tan φ
cos2 θ
π
2
Z
π
4
1
π
π
d sin θ
√
(e
)dθ = esin 2 − esin 4 = e − e 2
dθ
Question 13. State the Fundamental Theorem of Calculus.
Answer 13. (a) Let f be integrable on [a, b], and define a function F on [a, b] by
Z x
F (x) =
f (t)dt.
a
If f is continuous at c ∈ [a, b], then F is differentiable at c, and
dF
(c) = f (c).
dx
(b) If f is integrable on [a, b] and f (t) = dg
dt for some function g(t), then
Z b
f (t)dt = g(b) − g(a).
a
Question 14. Calculate
R8
du
.
2 u+1
Answer 14.
Z
8
2
Question 15. Calculate
du
8
= [log(u + 1)]2 = log(9) − log(3) = log(3)
u+1
R +√3
2
√ sin(x) e3x dx.
− 3 x4 +1
Answer 15. The integrand is an odd function of x, so
Z +√3
sin(x) 3x2
e dx = 0.
√
4+1
x
− 3