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Let f be a function such that 𝑓(−𝑥) = −𝑓(𝑥)  for all x. If ∫02𝑓(𝑥)𝑑𝑥=5, then ∫−22(𝑓(𝑥)+6)𝑑𝑥=Responses661616242434

Question

Let f be a function such that f(x)=f(x) f(-x) = -f(x) for all x. If 02f(x)dx=5 \int_0^2 f(x) \, dx = 5 , then 22(f(x)+6)dx= \int_{-2}^2 (f(x) + 6) \, dx =

Responses

661616242434

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Solution

The function f(x) is an odd function because it satisfies the property f(-x) = -f(x).

The integral of an odd function over a symmetric interval is zero. This is because the areas above and below the x-axis cancel each other out.

So, ∫ from -2 to 2 of f(x) dx = 0.

Now, consider the integral ∫ from -2 to 2 of (f(x) + 6) dx.

This can be split into two integrals: ∫ from -2 to 2 of f(x) dx + ∫ from -2 to 2 of 6 dx.

As we've already established, ∫ from -2 to 2 of f(x) dx = 0.

The integral ∫ from -2 to 2 of 6 dx is a simple one. The antiderivative of 6 is 6x, so this integral evaluates to 62 - 6(-2) = 24.

Therefore, ∫ from -2 to 2 of (f(x) + 6) dx = 0 + 24 = 24.

This problem has been solved

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