Step-by-step explanation:
∫₋₂² (f(x) + 6) dx
Split the integral:
∫₋₂² f(x) dx + ∫₋₂² 6 dx
Graphically, if f(-x) = -f(x), then ∫₋₂² f(x) dx = 0. But we can also show this algebraically.
Split the first integral:
∫₋₂⁰ f(x) dx + ∫₀² f(x) dx + ∫₋₂² 6 dx
Using substitution, write the first integral in terms of -x.
∫₂⁰ f(-x) d(-x) + ∫₀² f(x) dx + ∫₋₂² 6 dx
-∫₂⁰ f(-x) dx + ∫₀² f(x) dx + ∫₋₂² 6 dx
Flip the limits and multiply by -1.
∫₀² f(-x) dx + ∫₀² f(x) dx + ∫₋₂² 6 dx
Rewrite f(-x) as -f(x).
∫₀² -f(x) dx + ∫₀² f(x) dx + ∫₋₂² 6 dx
-∫₀² f(x) dx + ∫₀² f(x) dx + ∫₋₂² 6 dx
The integrals cancel out:
∫₋₂² 6 dx
Evaluating:
6x |₋₂²
6 (2 − (-2))
24
