You can use the formula for the error estimate for the trapezoidal rule.
∣ E T ∣ = ( b − a ) 3 K 2 12 n 2 \displaystyle |E_{T}|=\frac{(b-a)^{3}K_{2}}{12n^{2}} ∣ E T ∣ = 1 2 n 2 ( b − a ) 3 K 2
Where
K 2 \displaystyle K_{2} K 2 is the max value of the second derivative,
∣ f ′ ′ ( x ) ∣ \displaystyle |f''(x)| ∣ f ′ ′ ( x ) ∣ , on the interval [a,b].
You did not state, but your interval is [0,9], I believe. Because
∫ 0 9 [ 6 t + 40 ] d t = 468 \displaystyle \int_{0}^{9}[6\sqrt{t}+40]dt=468 ∫ 0 9 [ 6 t + 4 0 ] d t = 4 6 8 f ′ ′ ( x ) = − 3 2 t 3 2 \displaystyle f''(x)=\frac{-3}{2t^{\frac{3}{2}}} f ′ ′ ( x ) = 2 t 2 3 − 3 ∣ E T ∣ = ( 9 − 0 ) 3 ⋅ K 2 12 n 2 ≤ . 01 \displaystyle |E_{T}|=\frac{(9-0)^{3}\cdot K_{2}}{12n^{2}}\leq .01 ∣ E T ∣ = 1 2 n 2 ( 9 − 0 ) 3 ⋅ K 2 ≤ . 0 1
It is difficult to find a max value on [0,9], and inequalities are of little value in finding an upper bound on the magnitude of the
error. Because,
f ′ ′ ( x ) → − ∞ \displaystyle f''(x)\to {-\infty} f ′ ′ ( x ) → − ∞ as
t → 0 + \displaystyle t\to 0^{+} t → 0 +
So, in cases where it is rough to find values of
K 2 \displaystyle K_{2} K 2 , they may be replaced with any larger constant if the constant is easier to find.
For instance, if
K 2 < K \displaystyle K_{2}<K K 2 < K , then
∣ E T ∣ ≤ ( b − a ) 3 K 2 12 n 2 < ( b − 1 ) 3 K 12 n 2 \displaystyle |E_{T}|\leq \frac{(b-a)^{3}K^{2}}{12n^{2}}<\frac{(b-1)^{3}K}{12n^{2}} ∣ E T ∣ ≤ 1 2 n 2 ( b − a ) 3 K 2 < 1 2 n 2 ( b − 1 ) 3 K
So, the right side is also an upper bound on the value of
∣ E T ∣ \displaystyle |E_{T}| ∣ E T ∣ , though it is larger and not quite as good as using
K 2 \displaystyle K_{2} K 2 .
Try some values for K and solve for n.
