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.

draw a picture !

(a) 5*2 = 10
4 + 10 = 14

(b) 2 * 4 = 8

Unless I am missing something :)

(a) Well, let's start by looking at the original integral: ∫[0, 5] f(x) dx = 4.

Now, we want to evaluate the integral of f(x) + 2 over the same interval [0, 5].

Using the linearity property of the definite integral, we can split this into two separate integrals:
∫[0, 5] f(x) dx + ∫[0, 5] 2 dx.

But we know that the first integral is equal to 4, so we have:
4 + ∫[0, 5] 2 dx.

Now, integrating a constant function like 2 is pretty straightforward. The integral of 2 with respect to x is simply 2x.

Evaluating this integral over the interval [0, 5], we get:
4 + [2x] from 0 to 5.

Plugging in the upper and lower limits, we have:
4 + [2(5) - 2(0)].

Simplifying this expression, we obtain:
4 + 10 - 0 = 14.

So, the value of the integral ∫[0, 5] [f(x) + 2] dx is 14.

(b) Since f(x) is an even function, this means that the graph of f(x) is symmetric about the y-axis.

Now, the interval [-5, 5] is symmetric about the y-axis as well.

So, we can split the integral ∫[-5, 5] f(x) dx into two equal halves:

∫[0, 5] f(x) dx + ∫[-5, 0] f(x) dx.

But, since f(x) is an even function, it means that f(-x) = f(x).

So ∫[-5, 0] f(x) dx is just the same as ∫[0, 5] f(-x) dx.

Now we can rewrite the integral as:
∫[0, 5] f(x) dx + ∫[0, 5] f(-x) dx.

But the trick here is that these two integrals are actually the same because the function is even!

So, we can rewrite it as:
2∫[0, 5] f(x) dx.

And we already know from part (a) that ∫[0, 5] f(x) dx = 4.

Plugging this into the rewritten integral, we obtain:
2(4) = 8.

Therefore, the value of the integral ∫[-5, 5] f(x) dx, when f is an even function, is 8.

Hope that brings a smile to your face!

(a) To evaluate the definite integral from 0 to 5 of [f(x) + 2] dx, we can use the linearity property of the definite integral. According to this property, if we have a constant k, we can extract it from the integral and then integrate the function.

Therefore, we can rewrite the given integral as:

∫(0 to 5) [f(x) + 2] dx = ∫(0 to 5) f(x) dx + ∫(0 to 5) 2 dx

The first integral is already given as 4, so we have:

∫(0 to 5) [f(x) + 2] dx = 4 + 2∫(0 to 5) dx

Integrating the constant function 2 with respect to x gives us:

∫(0 to 5) [f(x) + 2] dx = 4 + 2[x] from 0 to 5

Evaluating the definite integral gives us:

∫(0 to 5) [f(x) + 2] dx = 4 + 2(5) - 2(0) = 14

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

(b) Given that f is an even function, we know that f(-x) = f(x) for all x in the interval [-5, 5]. Therefore, we can rewrite the definite integral from -5 to 5 of f(x) dx as the sum of two identical integrals over the interval [0, 5].

∫(-5 to 5) f(x) dx = ∫(-5 to 0) f(x) dx + ∫(0 to 5) f(x) dx

But we are given that ∫(0 to 5) f(x) dx = 4. Therefore, we have:

∫(-5 to 5) f(x) dx = ∫(-5 to 0) f(x) dx + 4

Since f is an even function, integrating from -5 to 0 is the same as integrating from 0 to 5 but with a negative sign:

∫(-5 to 5) f(x) dx = -∫(0 to 5) f(x) dx + 4

Substituting the value of the definite integral ∫(0 to 5) f(x) dx = 4, we have:

∫(-5 to 5) f(x) dx = -(4) + 4 = 0

Therefore, the value of the definite integral from -5 to 5 of f(x) dx, given that f is an even function, is 0.

To solve both parts of the question, we'll need to use the properties of the definite integral. Let's start with part (a).

(a) We want to evaluate the definite integral from 0 to 5 of [f(x) + 2] dx.

Using the linearity property of the definite integral, we can split this integral into two separate integrals:

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

We know that the definite integral from 0 to 5 of f(x) dx is equal to 4, as given in the question. So the first integral is simply 4.

The definite integral of a constant function such as 2 over any interval is equal to the product of the constant and the interval. In this case, the interval is from 0 to 5, so the second integral is (2 * 5) = 10.

Adding these two values together, we get 4 + 10 = 14. Therefore, the value of the definite integral from 0 to 5 of [f(x) + 2] dx is 14.

Now let's move on to part (b).

(b) We want to evaluate the definite integral from -5 to 5 of f(x) dx, given that f is an even function.

Since f is an even function, its graph is symmetric about the y-axis. This means that the definite integral from -5 to 5 of f(x) dx can be split into two equal halves:

Definite integral from -5 to 0 of f(x) dx + Definite integral from 0 to 5 of f(x) dx

Since these two integrals have the same limits but opposite signs, their sum cancels out to zero:

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

Therefore, the value of the definite integral from -5 to 5 of f(x) dx is zero.

To summarize:

(a) Definite integral from 0 to 5 of [f(x) + 2] dx = 14
(b) Definite integral from -5 to 5 of f(x) dx = 0