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3 years ago

Hey can anyone help me with this pls? :3 tysm in advance

Advanced Placement (AP)

2 answers:

Dmitry [639]3 years ago

8 0

Yeah there’s nothing there

Veseljchak [2.6K]3 years ago

4 0

With what? There is no picture or question added

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Read 2 more answers

I can’t figure this problem out

Natalka [10]

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

Lets first break up our integrand into two integrals

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

Now we can evaluate this:

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

So:

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

-15 + 2(3)

So we are left with -15 + 6 = -9

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3 years ago