Exercises: Integrals
Immediate integrals or Primitives
1. Determine 𝑓(𝑥) given that 𝑓′(𝑥)=12𝑥2−4𝑥 and 𝑓(−3)=17
Solution: 𝑑𝑓
𝑑𝑥=12𝑥2− 4𝑥 ⟹𝑑𝑓=(12𝑥2−4𝑥)𝑑𝑥 ⟹∫ 𝑑𝑓=∫(12𝑥2−4𝑥)𝑑𝑥=
12
3𝑥3− 4
2𝑥2+ 𝐶
Therefore: 𝑓(𝑥)=4𝑥3−2𝑥2+𝐶
Now, with initial conditions: 𝑓(−3)= 4 × (−3)3−2(−3)2+ 𝐶 ⟹17=−108−18+
𝐶 ⟹𝐶= 108+18+17 ⟹𝐶= 143 ⟹ 𝑓(𝑥)=4𝑥3−2𝑥2+143.
2. Determine 𝑔(𝑧) given that 𝑔′(𝑧)=3𝑧3+7
2√𝑧−𝑒𝑧 and 𝑔(1)=15−𝑒
Solution: 𝑑𝑔
𝑑𝑧=𝑔′(𝑧)=3𝑧3+7
2√𝑧−𝑒𝑧 ⟹ 𝑑𝑔=(3𝑧3+7
2√𝑧−𝑒𝑧)𝑑𝑧 ⟹∫ 𝑑𝑔=
∫(3𝑧3+7
2√𝑧−𝑒𝑧) 𝑑𝑧 ⟹𝑔(𝑧)=3
4𝑧4+7√𝑧−𝑒𝑧+𝐶
With Initial condition: 𝑔(1)=15−𝑒=3
4+ 7 − 𝑒 + 𝐶 ⟹ 𝐶=15−7−3
4=8−3
4=29
4
𝑔(𝑧)=3
4𝑧4+7√𝑧−𝑒𝑧+29
4
3. ∫(8𝑥−12)(4𝑥2−12𝑥)4𝑑𝑥
Solution: We use the power’s rule ∫𝑓𝑛(𝑥).𝑓′(𝑥)𝑑𝑥=𝑓𝑛+1 (𝑥)
𝑛+1 +𝐶
∫(8𝑥−12)(4𝑥2−12𝑥)4𝑑𝑥=(4𝑥2−12𝑥)5
5+𝐶
4. ∫3𝑡−4(2+4𝑡−3)−7𝑑𝑡
Solution:
∫3𝑡−4(2+4𝑡−3)−7𝑑𝑡=∫ −3(−12
12 𝑡−4(2+4𝑡−3)−7𝑑𝑥
= − 3
12∫ −12𝑡−4(2+4𝑡−3)−7𝑑𝑥=−1
4(2+4𝑡−3)−6
−6 +𝐶
=1
24(2+4𝑡−3)−6+𝐶
5. ∫5(𝑧−4)√𝑧2−8𝑧
3 𝑑𝑥
Solution:
