MATH1051 Lecture 17: Lecture 17
Lecture #17 – Midterm 2 Review
Example #1
●dt
∫
t5
ln(t)
○Notice that we can split this into two products
●n(t)(t)dt
∫
l−5
○Now that we have two values, notice that we can use integration by parts
● ln(t), therefore,du dtu = = t
1
●v t dt,therefore,v d= −5 = −4
t−4
●dt ln(t)( ) ( ) t
∫
t5
ln(t)= −4
t−4 − ∫
−4
t−4 (t
1)d
●ln(t)( ) ( ) C = −4
t−4 + 4
1−4
t−4 +
●t(ln(t)) t C = − 4
1−4 − 1
16
−4 +
Example #2
●sin(3x)dx
∫
ex
○Integration by parts again
○ sin(3x), therefore,du 3(cos(3x))dxu = =
○v e dx,therefore,v ed = x = x
●sin(3x)dx sin(3x)e (3(cos(3x)))dx
∫
ex= x− ∫
ex
●sin(3x)dx sin(3x)e 3 (cos(3x))dx
∫
ex= x− ∫
ex
○cos(3x)dx
∫
ex
■ cos(3x), therefore,du 3(sin(3x))dxu = =
■v e dx,therefore,v ed = x = x
■cos(3x) cos(3x)e 3 sin(3x)dx
∫
ex= x+ ∫
ex
●sin(3x)dx sin(3x)e 3(cos(3x)e3sin(3x)dx)
∫
ex= x− x+ ∫
ex
●sin(3x)dx sin(3x)e 3(cos(3x))e 9 sin(3x)dx
∫
ex= x− x− ∫
ex
●0sin(3x)dx sin(3x)e 3(cos(3x))e1∫
ex= x− x
●sin(3x)dx ( )sin(3x)e 3(cos(3x))e C
∫
ex= 1
10 x− x+
Example #3
●in θdθ in θ(sinθ)dθ
∫
s5= ∫
s4
●(1 cos θ) (sinθ)dθ
∫
− 22
○ cosθu=
○u sinθdθd=
●(1 ) du−∫
−u22
●−u du
∫
u4+ 2 2− 1
●u u u C−5
15+ 3
23− +
●cos θ cos θ cosθ C−5
15+ 3
23− +
Example #4
●dx
∫
4
2√2
1
x√x−4
2
○dx
∫
4
2√2
1
2x√−1
4
x2
●,so u , and therefore,du dxu2=4
x2 = x
2 = 2
1
Document Summary
Notice that we can split this into two products n(t)(t. Now that we have two values, notice that we can use integration by parts ln(t), therefore, du t dt, therefore, v u = d = 5 v t 4 + 4. = 4 t 4 ( t t 4 . Integration by parts again sin(3x), therefore, du. = x e e dx, therefore, v u = d = x v. 3(cos(3x))dx ex sin(3x)dx sin(3x)e x ex (3(cos(3x)))dx ex sin(3x)dx sin(3x)e x . 3 ex (cos(3x))dx ex cos(3x)dx u = d = x v cos(3x), therefore, du e dx, therefore, v. 3(sin(3x))dx ex sin(3x)dx ex cos(3x) cos(3x)e x + . 3 ex sin(3x)dx) ex sin(3x)dx sin(3x)e x . Example #3 (1 s 5 in d s 4 in (sin )d . 2 u = u d = cos sin d (1. ) du u4 + 2 2 1 du u. 1 x x 42 dx x2 u2 = 4.
