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Suppose $\int_{1}^{4} f(x) d x=8$ and $\int_{1}^{6} f(x) d x=5 .$ Evaluate the following integrals.a. $\int_{1}^{4}(-3 f(x)) d x$$\mathbf{b} . \int_{1}^{4} 3 f(x) d x$c. $\int_{6}^{4} 12 f(x) d x$$\mathbf{d .} \int_{4}^{6} 3 f(x) d x$

a. -24b. 24c. 36d. -9

Calculus 1 / AB

Chapter 5

Integration

Section 2

Definite Integrals

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Okay. This question gives us some integral and it wants us to use the linearity properties to find different modified forms. So part A, once the integral from 1 to 4 of negative, three times f of x, the X and we know that we can just factor out Constance. So this is just negative. Three times are integral from 1 to 4 of f of X, and we know that the value of this integral is just eight. So this integral turns out to just be 24 then for part B. It wants the integral from 1 to 4 of three times f of X, and this is the exact same as the last one. Except now are constant is a positive three. So we can just pull it out front. And we know that this integral eight. So it's just eight times three, which is 24. Done part C once the integral from 6 to 4 of 12 times after vax d x. So we don't really like our limits going this way so we could make it negative and pull out the 12 at the same time. So let's get negative 12 times the integral from 4 to 6 of F of x DX. Because again we just pull Constance out in front and to get our limits facing the direction we want, we just pull a negative sign. And now we actually don't have an expression for this integral. But we can figure out its value because we know that this whole integral has to be equal to the entire area minus the area from 1 to 4. So this just turns out speak negative 12 times five minus eight, which is just 12 times negative three or positive 36. And again, we found that into girl just by taking the total area from 1 to 6 and subtracting the area we already had at X equals four to get the area between. So then, lastly, Part D wants the integral from 4 to 6 of three f of x dx, and that's just equal to three times the integral from 4 to 6 of f of X. And we actually figured out that integral, but we can show it again. The integral from 4 to 6 again is just the integral going all the way from 1 to 6 minus the area we have accumulated up until X equals four and I'll change that limit cause I miss wrote it. So it's just three times negative three, which is negative and nine.

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