CALCULUS II

TUTORIAL 2 ALL COHORTS 2 SEPTEMBER 2022

TECHNIQUES OF INTEGRATION

1. THE FUNDAMENTAL THEOREM OF CALCULUS/POWER RULE

The Fundamental Theorem of Calculus tells us that the exact area under a continuous function f
over an interval [𝑎, 𝑏]is calculated directly using a definite integral:

𝑏
∫ f(𝑥)𝑑𝑥 = F(b) − F(a)
𝑎

The function is any antiderivative of (we usually set the constant of integration equal to 0).

1
Power Rule of Antidifferentiation: ∫ 𝑥 𝑛 𝑑𝑥 = 𝑛+1 𝑥 𝑛+1 + 𝑐, 𝑛 ≠ 1

Integrate the following:

 (x + 3) dx
2 4 2
 
2
1. i. x 3 dx ii. xdx iii. 2
0 0 −2

 1 1 
 ( 3x − 4 x + 2 ) dx
2 −1
 − 3  dx
2
2. i. ii.
−2  x 2
iii. ∫(5𝑥 + 7) 𝑑𝑥
−1
 x 

3. i. ∫(𝑥 + 1)2 𝑑𝑥 ii. ∫(𝑡 2 + 5𝑡 + 1)𝑑𝑡 iii. ∫(𝑥 + 1)2 𝑑𝑥

1 1⁄
4. i ∫ (𝑥 2 + 𝑥 2 ) 𝑑𝑥 ii. ∫(𝑡 3 + 𝑡 2 )𝑑𝑡 iii. ∫𝑡 2 𝑑𝑡

3 2
5. i. ∫ 3√𝑤 𝑑𝑤 ii. ∫(𝑥 3 + 5𝑥 2 + 6)𝑑𝑥 iii. ∫ ( 𝑡 − 𝑡 2 ) 𝑑𝑡

2 1 5 1
6. i. .∫1 𝑡2
𝑑𝑡 ii. ∫2 (𝑥 3 − 𝜋𝑥 2 )𝑑𝑥 iii. ∫0 (𝑦 2 + 𝑦 4 )𝑑𝑦

1 9 1
7. i. ∫0 2𝑒 𝑡 𝑑𝑡 ii. ∫4 √𝑥 𝑑𝑥 iii. ∫0 (6𝑞 2 + 4) 𝑑𝑞

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 5 4 1 2
8. i. ∫0 3𝑥 2 𝑑𝑥 ii. ∫1 𝑥
𝑑𝑡 iii. ∫0 (3𝑡 2 + 4𝑡 + 3)𝑑𝑡
√

1 21 2 1
9. i. ∫ (𝑦 + 𝑦 + 1) 𝑑𝑦 ii. ∫1 𝑥 𝑑𝑥 iii. ∫1 𝑥2
𝑑𝑥

1
10. i. ∫(4𝑡 + 7)𝑑𝑡 ii. ∫(𝑥 + 1)2 𝑑𝑥 iii. ∫ (𝑥 + √𝑥) 𝑑𝑥

2. INTEGRATION BY SUBSTITUTION

The following formulas provide a basis for an integration technique called substitution:

𝑢𝑟+1
A. ∫ 𝑢𝑟 𝑑𝑢 = 𝑟+1
+ 𝑐 , assuming 𝑟 ≠ −1

B. ∫ 𝑒 𝑢 𝑑𝑢 = 𝑒 𝑢 + 𝑐
1 1
C. ∫ 𝑢 𝑑𝑢 = ln|𝑢| + 𝑐, or ∫ 𝑢 𝑑𝑢 = ln⌈𝑢⌉ + 𝑐, 𝑢 > 0

Other identities:

−1
𝑎. ∫ sin(𝑎𝑥 + 𝑏)𝑑𝑥 = cos(𝑎𝑥 + 𝑏)
𝑎
+𝑐

1
𝑏. ∫ cos(𝑎𝑥 + 𝑏)𝑑𝑥 = sin(𝑎𝑥 + 𝑏) + 𝑐
𝑎

1
∫ 𝑑𝑥 = ln|𝑎𝑥 + 𝑏| + 𝑐
𝑎𝑥 + 𝑏

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Integrate the following using Substitution:

1
1. i. ∫(𝑥 + 4)5 𝑑𝑥 ii.∫ cos(3𝑥 + 4)𝑑𝑥 iii ∫ 1−2𝑥 𝑑𝑥

3 1
2. i. ∫1 (9 + 𝑥)2 𝑑𝑥 ii.∫(𝑥 − 2)3 𝑑𝑥 iii. ∫0 (𝑥 + 5)4 𝑑𝑥

1
3. i. ∫(2x − 1)7 dx ii.∫−1(1 − x)3 dx iii. ∫ sin(7x − 3)dx

1
4. i. ∫ 𝑒 3𝑥−2 𝑑𝑥 ii.∫ 7𝑥+5 𝑑𝑥 iii. ∫ 2𝑥√1 + 𝑥 2 𝑑𝑥

4𝑥 sin √𝑥
5. i. ∫ √2𝑥 2 𝑑𝑥 ii. ∫ 𝑑𝑥 iii. ∫ 5𝑥√1 − 𝑥 2 𝑑𝑥
+1 √𝑥

𝑑𝑥 𝑥3
6. i. ∫ 2 ii. ∫ 𝑥 4 (1 + 𝑥 5 )3 𝑑𝑥 iii. ∫ √𝑥 4 𝑑𝑥
√𝑥(1+√𝑥) +16

cos 𝑥 1 𝑥3
7. i. ∫ (5+sin 𝑥)2 𝑑𝑥 ii. ∫0 𝑑𝑥 iii. ∫ 5𝑥 2 √1 − 𝑥 3 𝑑𝑥
√𝑥 4 +12

5
8. i. ∫ 𝑒 cos 𝑥 sin 𝑥 𝑑𝑥 ii. ∫ 𝑒 𝑠𝑖𝑛𝑥 cos 𝑥 𝑑𝑥 iii. ∫0 𝑥 3 √𝑥 4 + 1𝑑𝑥

1 1 1 5
9. i. ∫0 𝑥(𝑥 2 + 1)5 dx ii. . ∫0 𝑥√𝑥 2 + 1dx iii. ∫0 (5𝑥+1)2
𝑑𝑥

2 𝑥+1 3 𝑑𝑥 1 5
10. i. ∫1 (𝑥 2 +2𝑥)2
𝑑𝑥 ii. ∫0 iii. ∫0 (5𝑥+1)2
𝑑𝑥
𝑥+1

3. INTEGRATION BY PARTS

Recall the Product Rule for differentiation,

𝑑 𝑑𝑣 𝑑𝑢
(𝑢𝑣) = 𝑢 +𝑣
𝑑𝑥 𝑑𝑥 𝑑𝑥

Integrating both sides with respect to 𝑥, we get

𝑑𝑣 𝑑𝑢
𝑢𝑣 = ∫ 𝑢 𝑑𝑥 + ∫ 𝑣 𝑑𝑥
𝑑𝑥 𝑑𝑥

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 = ∫ 𝑢𝑑𝑣 + ∫ 𝑣𝑑𝑢

Solving for ∫ 𝑢𝑑𝑣, we get the following theorem,

∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢

Use Integration by Parts to integrate the given functions

 x cos xdx x e e
2 −x x
1. i. ii. dx iii. sin xdx

 x e dx x 1 − xdx  x sin xdx

x  sin x
2 2 2
3. i. cos xdx ii. xdx iii. ln xdx

 xe x  ( ln x )
3x 3 2
4. i. dx ii. ln xdx iii. dx

(x + 1)dx  ( 5 + 3x ) e  x ( ln x )
2 −x 3
2
5. i. ii. dx iii. dx

 xe  te  (1 − x ) e dx
−x 1−t x
6. i. dx ii. dt iii.

 t ln 2tdt x x − 6dx  x ( ln x )
2
7. i. ii. iii. dx

3
8. i. ∫0 ln(𝑥 + 7)𝑑𝑥 ii. ∫ ln(𝑥 + 4)𝑑𝑥 iii. ∫ 𝑥 2 ln(5𝑥)𝑑𝑥

9. i. ∫ 𝑥 √𝑥 + 2𝑑𝑥 ii. ∫(𝑥 − 1) ln 𝑥 𝑑𝑥 iii. ∫ ln(𝑥)2 𝑑𝑥

10. i. ∫ 𝑥 ln √𝑥𝑑𝑥 ii. ∫ 𝑥𝑒 −𝑥 𝑑𝑥 iii. ∫ 𝑥 2 (2𝑥)𝑑𝑥

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