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dangina [55]
3 years ago
15

Astronauts in orbit feel weightless because

Advanced Placement (AP)
1 answer:
Nesterboy [21]3 years ago
8 0
There is no gravity and the air is denser
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3 years ago
I can’t figure this problem out
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Answer:

Explanation:

Alright so the way to do this is to use properties of integrals to make our life easier.

So we have:

\int\limits^4_1 {(3f(x)+2)} \, dx

So lets break this up into two different integrals that represent the same area.

\int\limits^4_0 {f(x)} \, dx - \int\limits^1_0 {f(x)} \, dx = \int\limits^4_1 {f(x)} \, dx

Lets think about what is going on up there. The integral from four to zero gives us the area under the curve of f(x) from four to zero. If we subtract this from the integral from one to zero (the area under f from one to zero) we are left with the area under f from four to one! Hence:

\int\limits^4_1 {f(x)} \, dx

But since we have these values we can say that:

-3 - 2 = -5

Which means that \int\limits^4_1 {f(x)} \, dx = -5

So now we can evaluate \int\limits^4_1 {(3f(x)+2)} \, dx

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\int\limits^4_1 {(3f(x)+2)} \, dx = 3\int\limits^4_1{f(x)} \, dx + 2\int\limits^4_1 {} \, dx

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We know that \int\limits^4_1 {f(x)} \, dx = -5

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3(-5)+2[x] where x is evaluated at 4 to 1 so

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