To solve this integral, we have to consider that
e3x+1=euu=3x+1,u(1)=4,u(0)=1⟹du=3dx
With that information we proceed and find:
I=∫0118e3x+1dx=6(∫013e3x+1dx)⟹I=6(∫14eudu)=6([eu]14)=6(e4−e1)
⟹I≊6(54.5981−2.7183)≊311.2788≊311
In conclusion, ∫10 18e3x+1dx (round to an integer) is equal to option 2. 311.
Reference:
- Varberg, D. E., Purcell, E. J., & Rigdon, S. E. (2007). Calculus with differential equations. Pearson/Prentice Hall.
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