1. ## Trapezoidal Rule

Can i please solution to this question , thanks

2. The four subdivisions are 1 -> 1.5, 1.5 -> 2, 2 -> 2.5 and 2.5 -> 3.

So you calculate the value of the function at each of the 5 points at the boundaries of these intervals and apply the rule for finding the area of a trapezoid:

A = x (y_1 + y_2) / 2

for each trapezoid, then add them all up.

3. Sorry, i still dont get it wats the forumula you are using

4. Take the first trapezoid. That is, the one that goes from x=1 to x=1.5.

Thus the x distance is 0.5.

The y distance at x=1 (which we can call y_1) is f(1) = 16-1=15 (from f(x) = 16-x^2).

The y distance at x=1.5 (which we can call y_2) is f(1.5) = 16-(1.5)^2 = 16-2.25 = 13.75.

So the area of the trapezoid = x (y_1 + y_2) / 2 (that is, multiply x by the mean of y_1 and y_2) is 0.5 (16 + 13.75) / 2 = whatever your calculator gives you.

Do the same thing with all the other intervals between 1.5 and 2.

Then take all these 4 areas and add them up.

5. Originally Posted by Misa-Campo

Can i please solution to this question , thanks
The integration is from x=1 to x=3. Four sub-intervals are needed, so
(3 -1)/ 4 = 0.5 per interval.

We need to find the y-values when x=1, x=1.5, x=2, x=2.5, and x=3
y = f(x) = 16 -x^2
So,
y(1) = 16 -1^2 = 15
y(1.5) = 16 -(1.5)^2 = 13.75
y(2) = 16 -2^2 = 12
y(2.5) = 16 -(2.5)^2 = 9.75
y(3) = 16 -3^3 = 7

So by trapezoidal rule,
INT(1 to 3)[16 -x^2]dx
= (d/2)[h1 +2h2 +2h3 +2h4 +h5]
= (0.5 /2)[15 +2(13.75) +2(12) +2(9.75) +7]
= 23.25 sq.units.

-------------
Check the exact integral:
INT.(1 to 3)[16 -x^2]dx
= [16x -(x^3)/3] |(1 to 3)
= [16(3) -(3^3)/3] -[16(1) -(1^3)/3]
= [48 -9] -[16 -1/3]
= 23.33 sq.units ------close to the approx. 23.25 sq.units.

6. ## x^3-6*x^2-x+30

Originally Posted by Misa-Campo

Can i please solution to this question , thanks