# Thread: Trapezoidal Rule Help

1. ## Trapezoidal Rule Help

1. Let R be the region in the first quadrant under the graph of $y=x(x^2+2)^3$ for x is greater than or equal to 0 and less and or equal to 2.

If the line x=k divides R into two regions of equal area, what is the value of k to the nearest .001?

I already know that the area of all of R=160. Do I have to integrate from 0 to k set to equal 80?(since the area is divided into two).

2. Let F(x)= (integral sign from 0 to x) sin (t^2)dt for x is greater than or equal to 0 and less than or equal to 3.

Use the trapezoid rule with four equal subdivisions of the closed interval [0,1] to approximate F(1).

It says that I need 4 equal rectangles, meaning I need 5 x and y values. My x values were 0, 1/4, 1/2, 3/4, and 1. But how do I use the given information to find y?

2. Originally Posted by DarkestEvil
1. Let R be the region in the first quadrant under the graph of $y=x(x^2+2)^3$ for x is greater than or equal to 0 and less and or equal to 2.

If the line x=k divides R into two regions of equal area, what is the value of k to the nearest .001?

I already know that the area of all of R=160. Do I have to integrate from 0 to k set to equal 80?(since the area is divided into two).

yes

2. Let F(x)= (integral sign from 0 to x) sin (t^2)dt for x is greater than or equal to 0 and less than or equal to 3.

Use the trapezoid rule with four equal subdivisions of the closed interval [0,1] to approximate F(1).

It says that I need 4 equal rectangles, meaning I need 5 x and y values. My x values were 0, 1/4, 1/2, 3/4, and 1. But how do I use the given information to find y?

$\textcolor{red}{F(1) = \int_0^1 \sin(t^2) \, dt \approx \frac{1}{8}\left[\sin(0) + 2\sin\left(\frac{1}{16}\right) + 2\sin\left(\frac{1}{4}\right)+2\sin\left(\frac{9}{ 16}\right)+\sin\left(1\right)\right]}$

...

### let f(x)= integral from 0 to x sin(t^2)dt

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