Let integral f(x)dx=9 from -6 to -1.5
integral f(x)dx=7 from -6 to -4.5
integral f(x)dx=4 from -3 to -1.5
A. Find integral f(x)dx from -4.5 to -3
B. Find integral (9f(x)-7)dx from from -4.5 to -3
you know that
∫[a,b] f(x) dx + ∫[b,c] f(x) dx = ∫[a,c] f(x) dx
That is adjoining intervals can be combined into one longer interval.
So,
∫[-6,-1.5] = ∫[-6,-4.5] + ∫[-4.5,-3] + ∫[-3,-1.5]
or,
9 = 7 + ∫[-4.5,-3] + 4
∫[-4.5,-3] = -2
and so on.
For the second, recall that
∫(a*f(x) + ∫g(x)) dx
= a∫f(x) dx + ∫g(x) dx
so, we have
∫[-4.5,3] 9f(x)-7 dx
= 9∫[-4.5,-3] f(x) dx - ∫[-4.5,-3] 7 dx
= 9(-2) - 7(-3-(-4.5))
= -28.5
A. To find the integral of f(x)dx from -4.5 to -3, we can use the integral properties:
∫[a to b] f(x)dx = ∫[a to c] f(x)dx + ∫[c to b] f(x)dx
Therefore, we can break down the integral from -4.5 to -3 into the sum of two integrals:
∫[-4.5 to -3] f(x)dx = ∫[-4.5 to -1.5] f(x)dx + ∫[-1.5 to -3] f(x)dx
From the given information, we know that:
∫[-4.5 to -1.5] f(x)dx = 7
∫[-1.5 to -3] f(x)dx = 4
So, the integral of f(x)dx from -4.5 to -3 is:
∫[-4.5 to -3] f(x)dx = 7 + 4 = 11
B. To find the integral of (9f(x) - 7)dx from -4.5 to -3, we can again use the integral properties:
∫[a to b] (f(x) + g(x))dx = ∫[a to b] f(x)dx + ∫[a to b] g(x)dx
Therefore, we can express the integral as the sum of two integrals:
∫[-4.5 to -3] (9f(x) - 7)dx = 9∫[-4.5 to -3] f(x)dx - 7∫[-4.5 to -3] dx
From the previous calculations, we know that:
∫[-4.5 to -3] f(x)dx = 11
Also, the integral of dx from -4.5 to -3 is simply the difference between the limits:
∫[-4.5 to -3] dx = -3 - (-4.5) = 1.5
So, the integral of (9f(x) - 7)dx from -4.5 to -3 is:
∫[-4.5 to -3] (9f(x) - 7)dx = 9(11) - 7(1.5) = 99 - 10.5 = 88.5
To solve the given problem, we need to use the properties of definite integrals.
A. To find the integral of f(x)dx from -4.5 to -3, we can subtract the integral of f(x)dx from -6 to -4.5 from the integral of f(x)dx from -6 to -3. This can be expressed as:
integral f(x)dx from -4.5 to -3 = integral f(x)dx from -6 to -3 - integral f(x)dx from -6 to -4.5
So, we need to find the values of the integrals from -6 to -3 and from -6 to -4.5. Using the given information, we know that the integral of f(x)dx from -6 to -3 is 9 and the integral of f(x)dx from -6 to -4.5 is 7. Plugging these values into the equation:
integral f(x)dx from -4.5 to -3 = 9 - 7 = 2
Therefore, the integral of f(x)dx from -4.5 to -3 is 2.
B. To find the integral of (9f(x)-7)dx from -4.5 to -3, we can use the linearity property of integrals. The linearity property states that ∫(af(x) + bg(x))dx = a∫f(x)dx + b∫g(x)dx, where a and b are constants.
So, we can break down the given integral into two separate integrals:
∫(9f(x)-7)dx from -4.5 to -3 = 9∫f(x)dx from -4.5 to -3 - 7∫dx from -4.5 to -3
Using the information provided, we know that ∫f(x)dx from -4.5 to -3 is 2 (as calculated in part A). The integral of dx is simply the difference of the bounds, -3 - (-4.5) = 1.5.
Plugging these values into the equation, we get:
∫(9f(x)-7)dx from -4.5 to -3 = 9(2) - 7(1.5) = 18 - 10.5 = 7.5
Therefore, the integral of (9f(x)-7)dx from -4.5 to -3 is 7.5.
Note: If more accurate values or information about f(x) are given, the specific values of the integrals will differ, but the general method for solving the problem remains the same.