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.
This was done by Damon, but here's my take
∫[0,5](f(x)+2) dx
= ∫[0,5] f(x) dx + ∫[0,5] 2 dx
= 4 + 2x[0,5]
= 4 + 10
= 14
Since we are dealing with an even function,
∫[-5,5] f(x) dx = 2∫[0,5] f(x) dx = 2*4 = 8
(a) Alright, let's evaluate the definite integral from 0 to 5 of [f(x) + 2]dx!
Now, we know that the integral of a constant multiplied by a function is equal to the constant multiplied by the integral of the function. In this case, we have a constant of 2 added to f(x), so the integral of 2 times f(x) will be 2 times the integral of f(x).
Since the definite integral from 0 to 5 of f(x)dx is equal to 4, we can write the definite integral from 0 to 5 of [f(x) + 2]dx as the integral from 0 to 5 of f(x)dx plus the integral from 0 to 5 of 2dx.
The integral from 0 to 5 of f(x)dx is 4, and the integral from 0 to 5 of 2dx is equal to 2 times the integral from 0 to 5 of dx, which is just the length of the interval from 0 to 5, which is 5 - 0 = 5.
So the integral from 0 to 5 of [f(x) + 2]dx is equal to 4 + 2 * 5 = 14.
(b) Alright, now let's evaluate the definite integral from -5 to 5 of f(x)dx, given that f is an even function.
We know that if f(x) is an even function, then f(x) = f(-x) for all x in the interval.
So, let's consider the interval from -5 to 5. Since f is an even function, we know that f(x) = f(-x) for every value of x in this interval.
This means that the definite integral from -5 to 5 of f(x)dx is the same as the definite integral from -5 to 5 of f(-x)dx.
Now, let's make a change of variable. Let u = -x. Then, dx = -du.
When x is -5, u is 5. When x is 5, u is -5.
So, the definite integral from -5 to 5 of f(-x)dx is equal to the definite integral from 5 to -5 of f(u)(-du) [after substituting -x by u and dx by -du].
Since the limits of integration have flipped, we can rewrite the integral as the definite integral from -5 to 5 of -f(u)du.
Now, we know that the integral of a function times -1 is equal to the negative of the integral of the function.
Therefore, the definite integral from -5 to 5 of -f(u)du is the negative of the definite integral from -5 to 5 of f(u)du.
But, since the definite integral from -5 to 5 of f(x)dx is given as 4, the definite integral from -5 to 5 of f(u)du is also equal to 4.
Thus, the definite integral from -5 to 5 of f(x)dx is also equal to 4.
I hope that answers your question! Remember, keep calm and calculate on! 😄
To evaluate each integral, we can use the properties of the definite integral.
(a) definite integral from 0 to 5 of [f(x) + 2] dx:
According to the properties of the definite integral, we can split the integral into two separate integrals:
definite integral from 0 to 5 of f(x) dx + definite integral from 0 to 5 of 2 dx
Since the definite integral from 0 to 5 of f(x) dx is given to be 4, we can substitute this value:
4 + definite integral from 0 to 5 of 2 dx
Integrating the constant term, we get:
4 + [2x] evaluated from 0 to 5
Plugging in the upper and lower limits of integration:
4 + [2(5) - 2(0)]
4 + 10 - 0
= 14
Therefore, the value of the definite integral from 0 to 5 of [f(x) + 2] dx is 14.
(b) definite integral from -5 to 5 of f(x) dx, given that f is an even function:
If f(x) is an even function, it means that f(-x) = f(x) for any x in the domain of f.
Using the property of even functions, we can rewrite the integral as:
2 times the definite integral from 0 to 5 of f(x) dx
Since the integral from 0 to 5 of f(x) dx is given to be 4, we can substitute this value:
2 * 4 = 8
Therefore, the value of the definite integral from -5 to 5 of f(x) dx is 8.
To evaluate the given definite integrals, we can use the properties of the definite integral. Here's how to approach each integral:
(a) The property we need to use here is the additive property of the definite integral. According to this property, the integral of the sum of two functions is equal to the sum of their individual integrals. Therefore, we can split the integral into two separate integrals:
∫[0 to 5] (f(x) + 2) dx = ∫[0 to 5] f(x) dx + ∫[0 to 5] 2 dx
We know that ∫[0 to 5] f(x) dx = 4 (given in the problem), and ∫[0 to 5] 2 dx is simply the integral of a constant, which results in the constant multiplied by the width of the interval:
∫[0 to 5] 2 dx = 2 * (5 - 0) = 10
Now, we can substitute these values back into the equation:
∫[0 to 5] (f(x)+2) dx = ∫[0 to 5] f(x) dx + ∫[0 to 5] 2 dx
= 4 + 10
= 14
Therefore, the value of the definite integral is 14.
(b) The property we need to use here is the symmetry property of the definite integral. According to this property, if the function is even, the integral from -a to a is equal to twice the integral from 0 to a.
Since f is an even function, we have:
∫[-5 to 5] f(x) dx = 2 * ∫[0 to 5] f(x) dx
And we know that ∫[0 to 5] f(x) dx = 4 (given in the problem).
Substituting this value into the equation, we get:
∫[-5 to 5] f(x) dx = 2 * ∫[0 to 5] f(x) dx
= 2 * 4
= 8
Therefore, the value of the definite integral is 8.
I hope this helps! Let me know if you have any further questions.