# simple trapizum rule problem

#### robertgreen

hi. this is a uncommon question on A level WJEC exam board and is an add on to a simple trapizium rule question (pure C2). maybe the simpson rule also(c3). it goes like this - use your answer for part (a) to deduce and approximate value for ..... (usually there is a sign change)

these questions are one mark so i know its something simple but it is rare so ive never been tought it. hahah feel embarrased asking. thanks

#### undefined

MHF Hall of Honor
hi. this is a uncommon question on A level WJEC exam board and is an add on to a simple trapizium rule question (pure C2). maybe the simpson rule also(c3). it goes like this - use your answer for part (a) to deduce and approximate value for ..... (usually there is a sign change)

these questions are one mark so i know its something simple but it is rare so ive never been tought it. hahah feel embarrased asking. thanks
Basically, you wrote down an equation in part (a), and you need to plug in numbers, right? Maybe if you gave a specific example and showed some work we could help better. If the subinterval length is not given, you might have to just choose one that seems appropriate or easy in context.

#### robertgreen

Basically, you wrote down an equation in part (a), and you need to plug in numbers, right? Maybe if you gave a specific example and showed some work we could help better. If the subinterval length is not given, you might have to just choose one that seems appropriate or easy in context.
here is an example from and old paper i've found-

(a) use the trapezium rule with five ordinates to find an approximate value for

integrate[log(1+X^2)(0 min)(1 max)]

answer im pretty sure is 0.09

(b) use your answer to part a to deduce an appropriate an approximate value for

integrate[log(sqrt(1+X^2))(0 min)(1 max)]

any workings needed i have

#### undefined

MHF Hall of Honor
here is an example from and old paper i've found-

(a) use the trapezium rule with five ordinates to find an approximate value for

integrate[log(1+X^2)(0 min)(1 max)]

answer im pretty sure is 0.09

(b) use your answer to part a to deduce an appropriate an approximate value for

integrate[log(sqrt(1+X^2))(0 min)(1 max)]

any workings needed i have
Ah, I didn't realize the integrand changed so much.

I was stumped by this for a bit but then realized this is a simple application of logarithm rules.

$$\displaystyle \int_0^1\log(\sqrt{x^2+1})\,dx$$

$$\displaystyle =\int_0^1\log\left((x^2+1)^{\frac{1}{2}}\right)\,dx$$

$$\displaystyle =\int_0^1\frac{1}{2}\cdot\log(x^2+1)\,dx$$

$$\displaystyle =\frac{1}{2}\cdot\int_0^1\log(x^2+1)\,dx$$

By the way, assuming log means natural log, I get 0.27 as my approximation for part (a).

• robertgreen

#### robertgreen

ahhhh. thanks for that pal. much appreciated