1 to 9$\displaystyle \int f(x)~dx$ = -1, 7 to 9$\displaystyle \int f(x)~dx$ = 5, 7 to 9$\displaystyle \int h(x)~dx$ = 4

How do you do this one?

(e) 7 to 1$\displaystyle \int f(x)~dx$

Printable View

- Jan 20th 2009, 06:44 PMkatchat64Definite Integrals & Antiderivatives
1 to 9$\displaystyle \int f(x)~dx$ = -1, 7 to 9$\displaystyle \int f(x)~dx$ = 5, 7 to 9$\displaystyle \int h(x)~dx$ = 4

How do you do this one?

(e) 7 to 1$\displaystyle \int f(x)~dx$ - Jan 20th 2009, 06:50 PMMush
The first integral finds the area under the curve on the interval $\displaystyle x = [1,9]$. The second integral finds the area under the curve on the interval $\displaystyle x = [-1,7]$. You want to find the area under the curve on the interval $\displaystyle x = [7,1] $. This is area is related to the other two areas, in that the area you want to find can be thought of as an area which intersects the first 2 areas!

- Jan 20th 2009, 06:58 PMkatchat64
Ok... I understand that part but my question is how do you know which ones to add or subtract to get 1 to 7 $\displaystyle \int f(x)~dx$

- Jan 20th 2009, 07:03 PMMush
Let's imagine that $\displaystyle f(x) = 2 $. I've attached a diagram to help you visualise the problem better.

The function is given by the blue line. The limits of the first integral are given by the green lines. The limits of the 2nd integral are given by the red lines. And the limits of the integral we're interested in happen to be involved in the first two!

Can you see it now? - Jan 20th 2009, 07:11 PMkatchat64
Ok... Now I understand it... the diagram really helped.... thanks! (Bow)