ap calc is hard!!!!....?

Question

if integral o to 4 f(x) dx= 10, integral 0 to 5 f(x) = 9 and integral 4 to 7 f(x)dx=1 then integra 5 to 7 f(x) dx =

Answer 1

Let F(x) be the antiderivitive of f(x)

F(4) - F(0) = 10
F(5) - F(0) = 9

Therefore F(5) - F(4) = -1

Also F(7) - F(4) = 1

So F(7) - F(5) = 1 - (-1) = 2

Answer 2

integral 0 to 5 - integral 0 to 4 = integral 4 to 5
So integral 4 to 5 = 9 -10 = -1
integral 5 to 7 = integral 4 to 7 - integral 4 to 5.
So integral 5 to 7 = 1 - (-1) = 2

Answer 3

[integral 5 to 7] =
[integral 0 to 7] minus [integral 0 to 5] =
[integral 0 to 4] plus [integral 4 to 7] minus [integral 0 to 5] =
10 + 1 - 9 = 2

Answer 4

Int {0 to 4} f(x)dx = 10
Int {0 to 5} f(x)dx = 9
Int {4 to 7} f(x)dx = 1

Find Int {5 to 7} from
Int {4 to 7} - Int {4 to 5}
= 1 - Int {4 to 5}

You don't have Int {4 to 5}, but you find it from
Int {0 to 5} - Int {0 to 4} = Int {4 to 5} = 9 - 10 = -1

Int {5 to 7} = = 1 - Int {4 to 5} = 1 - (-1) = 2

Answer 5

1) F(4)-F(0)=10
2) F(5)-F(0)=9
3) F(7)-F(4)=1

where F(x)'=f(x)

And you need F(7)-F(5) which you can get by subtracting 2 from 3 and adding 1to the result (i mean the equations).
You get:
F(7)-F(4)-F(5)+F(0)+F(4)-F(0)=1-9+10
Which means F(7)-F(5)=2

Answer 6

it relies upon on which calc you're taking calc AB basically counts for a semester worth of calculus the place as BC counts for a complete 12 months of faculty point calculus. with the intention to take those classes, you may desire to be a expert at integrating and deriving countless styles of applications, because of the fact in all honesty, this is all AP calc is, utilized derivations and integrations. in case you may no longer try this, you may no longer be taking AP calc.

Answer 7

graph the area under the function. it will come out into a shape. it will work, trust me, i did it.

if you cannot do it yourself, then you should really not be in ap calc