If we know that the definite integral from -6 to -3 of f(x) equals 6, the definite integral from -6 to -5 equals 2 and the definite integral from -4 to -3 equals 4 then:
What is the definite integral from -5 to -4? I know that this is zero.
But then what can we say about the definite integral from -4 to -5 of not just f(x) but 6f(x)-2?
i'm lost! I thought it would just be zero... Thank you!
Nevermind guys I got It!!!
To find the definite integral from -5 to -4, you can use the properties of integrals and the given information:
1. Recall that the definite integral represents the signed area under the curve. When integrating from a to b, where b is greater than a, the integral is positive if the curve lies above the x-axis between a and b and negative if the curve lies below the x-axis between a and b.
2. Given that the definite integral from -6 to -5 equals 2 and the definite integral from -4 to -3 equals 4, we can use these values to determine the behavior of the function between those intervals.
3. Since the definite integral from -6 to -3 is 6, the combined area between -6 to -5 and -5 to -4 should equal the total area from -6 to -3, which is 6.
4. Since the definite integral from -6 to -5 is 2, it means that the function f(x) is negative or zero between -6 and -5. Similarly, since the definite integral from -4 to -3 is 4, it means the function f(x) is positive or zero between -4 and -3.
5. Now, we know that the definite integral from -5 to -4 is zero. This indicates that the areas above the x-axis and below the x-axis cancel each other out, resulting in a net area of zero.
Moving on to the second part of your question, we need to determine what we can say about the definite integral from -4 to -5 of the function 6f(x) - 2.
Given that the definite integral from -5 to -4 of f(x) is zero (as we found earlier), we can write the new integral as the sum of two definite integrals:
∫[-4 to -5] 6f(x) - 2 dx = ∫[-4 to -5] 6f(x) dx - ∫[-4 to -5] 2 dx
Since the definite integral from -5 to -4 of f(x) is zero, the first term on the right-hand side of the equation becomes zero. Thus, we are left with:
∫[-4 to -5] 6f(x) - 2 dx = - ∫[-4 to -5] 2 dx = -2 ∫[-4 to -5] dx
Using the definition of the definite integral, we can simplify this further:
-2 ∫[-4 to -5] dx = -2 [x] [-4 to -5] = -2(-5 - (-4)) = -2(-5 + 4) = -2(-1) = 2
Therefore, we can conclude that the definite integral from -4 to -5 of the function 6f(x) - 2 is equal to 2.