consider the function f that is continuous on the interval [-5,5] and for which the definite integral 0(bottom of integral sign) to 5(top of integral sign) of f(x)dx=4. Use the properties of the definite integral to evaluate each integral:

(a) definite integral from 0(on the bottom of the integral sign) to 5(top of integral sign) of[f(x)+2]dx

(b) definite integral from -5(bottom of integral sign) to 5(top of integral sign) f(x)dx, given that f is an even function.

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To evaluate the integrals, we can use the properties of the definite integral.

(a) The definite integral of the function f(x) from 0 to 5 is given as 4. We need to find the definite integral of [f(x) + 2] from 0 to 5.

We can use the linearity property of the definite integral, which states that the integral of a sum of two functions is equal to the sum of their individual integrals.

Definite integral from 0 to 5 of [f(x) + 2]dx = Definite integral from 0 to 5 of f(x)dx + Definite integral from 0 to 5 of 2dx

We already know the definite integral from 0 to 5 of f(x)dx is 4.

For the integral of 2dx, the integral of a constant is equal to the constant times the interval of integration. Here, the constant is 2 and the interval of integration is from 0 to 5.

Definite integral from 0 to 5 of 2dx = 2 * (5 - 0) = 2 * 5 = 10

Therefore, the definite integral from 0 to 5 of [f(x) + 2]dx is equal to 4 + 10 = 14.

(b) Given that f is an even function, it means f(x) = f(-x) for all x in the domain.

The definite integral from -5 to 5 of f(x)dx represents the area under the curve of f(x) from -5 to 5. Since f is an even function, the area under the curve from -5 to 0 will be symmetric to the area under the curve from 0 to 5.

Therefore, the definite integral from -5 to 5 of f(x)dx will be equal to twice the definite integral from 0 to 5 of f(x)dx.

Definite integral from -5 to 5 of f(x)dx = 2 * (Definite integral from 0 to 5 of f(x)dx)

We are given that the definite integral from 0 to 5 of f(x)dx is 4. Plugging this value into the equation:

Definite integral from -5 to 5 of f(x)dx = 2 * 4 = 8

Therefore, the definite integral from -5 to 5 of f(x)dx, given that f is an even function, is equal to 8.