Use the Trapezoidal Rule and Simpson's Rule to approximate the value of the definite integral for the given value of n. Round your answer to four decimal places and compare the results with the exact value of the definite integral. http://tinypic.com/r/2ldwvoi/8
?? i really dont know this sorry but good luck \(\large\cal\color{red}=\color{red})\)
Thanks @Muzzack . :)
my pleasure :)
\[\large \int \limits_0^8 \sqrt[3]{x}~dx = \dfrac{\Delta x}{2}\left[f(1) + 2f(2) + \cdots + 2f(7) +f(8)\right]\]
\(\Delta x = \dfrac{b-a}{n} = \dfrac{8-0}{8} = 1\)
\[\large \int \limits_0^8 \sqrt[3]{x}~dx \approx \dfrac{1}{2}\left[f(1) + 2f(2) + \cdots + 2f(7) +f(8)\right]\]
you just need to evaluate the values and simplify @Lena772 ^
Thanks.. Do find the exact value i graph it?
To*
for exact value - as the question suggests, we have to evaluate the definite integral
you did the simpson's rule already ?
yes
Great ! use below formula for exact value : \[\large \int x^n~dx = \dfrac{x^{n+1}}{n+1}+C\]
\[\large \int \limits_0^8 \sqrt[3]{x}~dx = \int \limits_0^8 x^{\frac{1}{3}}~dx = ? \]
43.3137
is what I got for Simpsons Rule
that doesn't look right
what about trapezoidal ?
I got 0 for trapezoidal :|
All the numbers are positive in the sum, there is no way for that to become a 0 - check your calculation once :)
ok ... i now get -45.1687 for simpsons rule
try again, there is no way for that sum to go negative...
I get 12 for exact value...
Is that right @ganeshie8
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