- Evaluate the definite integral
∫−22x2dx=2∫02x2dx=2limϵ→0+∫ϵ2x2dx=2limϵ→0+(−x1)∣ϵ2=+∞
The first equality holds because the function x21,x∈[−2;2] is even.
The rest follows from improper integral definition and the fundamental theorem of calculus.
Answer: +∞ .
2. Evaluate the definite integral
∫−∞02xdx=∫−∞0eln(2)xdx=limΔ→−∞∫Δ0eln(2)xdx=limΔ→−∞(ln(2)eln(2)x)∣Δ0=ln(2)1
Answer: ln(2)1
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