In some serious need of intgral help?

Question

if you have ∫ -9 to -3 f(x)dx=5, ∫ -9 to -7 f(x)dx=2, and ∫ -5 to -3 f(x)dx =10, what is ∫ -5 to -7 (5f(x)-2)dx???

Answer 1

We use the fundamental law of calculus.

.. b
If ∫f(x)dx = F(b) - F(a) then we have...
. a

F(-3) - F(-9) = 5
F(-7) - F(-9) = 2
F(-3) - F(-5) = 10

F(-5) - F(-9) = [F(-3) - F(-9)] - [F(-3) - F(-5)] = 5 - 10 = -5

F(-5) - F(-7) = [F(-5) - F(-9)] - [F(-7) - F(-9)] = -5 - 2 = -7

We want to solve for S where

-5
∫[5f(x)-2]dx = S
-7

.-5 .......... -5
5∫f(x)dx - 2∫dx = S
.-7 ......... -7

S = 5[F(-5) - F(-7)] - 2[-5 - (-7)]
S = 5(-7) - 2(2)

S = -39

Answer 2

[-9, -3]∫f(x) dx = 5, [-9, -7]∫f(x) dx = 2, [-5, -3]∫f(x) dx = 10

Note that:

[-9, -3]∫f(x) dx = [-9, -7]∫f(x) dx + [-7, -5]∫f(x) dx + [-5, -3]∫f(x) dx

So:

5 = 2 + [-7, -5]∫f(x) dx + 10
-7 = [-7, -5]∫f(x) dx
7 = -[-7, -5]∫f(x) dx = [-5, -7]∫f(x) dx
35 = 5*[-5, -7]∫f(x) dx = [-5, -7]∫5f(x) dx
39 = [-5, -7]∫5f(x) dx + 4 = [-5, -7]∫5f(x) dx + [-5, -7]∫-2 dx
39 = [-5, -7]∫5f(x) - 2 dx

And we are done.

Edit: corrected careless arithmetic error.

Answer 3

integral from -9 to -3 =
integral from -9 to -7 +
integral from -7 to -5 +
integral from -5 to -3

integral from -9 to -3 = 5
integral from -9 to -7 = 2
integral from -7 to -5 = M (to be found)
integral from -5 to -3 = 10

5 = 2 +M +10 = 12 +M
M = 5 -12 = -7
we found the integral from -7 to -5 = -7
we want the integral from -5 to -7 it is the minus of the integral from -7 to -5
(integral = 7)
**********************************************
integral of {5f(x) -2}dx from -5 to -7 = 5(7) - (2)(-2)= 39

Answer 4

Ans:35