Integrals and Derivatives

• June 13th 2007, 04:17 PM
sw3etazngyrl
Integrals and Derivatives
Use the values to estimate integral from 0 to 6 using six equal subintervals with left endpoints.
(0, 9.7) (1, 8.9) (2, 8.4) (3, 6.1) (4, 4.5) (5, -6.9) (6, -10.3)

Find the interval on which the curve f(x)=integral of 1/(6+t+t^2)dt from 0 to x is concave downward.
• June 13th 2007, 09:07 PM
CaptainBlack
Quote:

Originally Posted by sw3etazngyrl
Use the values to estimate integral from 0 to 6 using six equal subintervals with left endpoints.
(0, 9.7) (1, 8.9) (2, 8.4) (3, 6.1) (4, 4.5) (5, -6.9) (6, -10.3)

You are expected to use a numerical integration rule to do this. The use of
the term "left end points" seems to imply that you are supposed to something
like:

$
I = \int_0^6 f(x) dx = \sum_{x=0}^5 f(x) \delta x = (9.7+8.9+8.4+6.1+4.5-6.9)\times 1 = 30.7
$

(note the last data point (6,-10.3) does not contribute to this as it is not the
left end point of one of the sub-intervals of (0,1). Simpsons rule would be a better
way of doing this if you know that).

RonL
• June 13th 2007, 09:11 PM
CaptainBlack
Quote:

Originally Posted by sw3etazngyrl

Find the interval on which the curve f(x)=integral of 1/(6+t+t^2)dt from 0 to x is concave downward.

Concave downwards means that the derivative is decreasing.

Now if:

$
f(x)=\int_0^x 1/(6+t+t^2)dt
$

by the fundamental theorem of calculus:

$
\frac{d}{dx} f(x)= 1/(6+x+x^2)
$

So can you take it from there?

RonL