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I have the question:

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For part (a) I got 10.449 and 3.357 enter image description here

For part (b) I got 8.68 with the following working:

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Is my answer for part (b) correct ? And if not could you point out at which point I went wrong ?

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    As a good check:$$\int_4^5\sqrt{x^3+2}\ dx=9.66565469368$$which means you did something wrong (likely missed something that came out to be $\approx+1$), as the trapezoidal sum should over-estimate, not under-estimate. – 2017-02-01
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    We may also notice that the Cauchy-Schwarz inequality provides an extremely accurate approximation: $$\frac{2}{5} \sqrt{3739-1411 \sqrt{5}}=\color{green}{9.6656}76\ldots$$ – 2017-02-01
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    Why the votes to close as not having enough context? As I see it, there is plenty of context here. – 2017-02-02
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    Yh I have tried to include as much context as possible even showed all my working out (: – 2017-02-02

2 Answers 2

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You made an error in part a). Notice that

$$f(5)=\sqrt{2+5^3}=\sqrt{127}>\sqrt{121}=11$$

So it must be something slightly larger than $11$. Other than that, the set-up everywhere looks fine.

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    Thanks (: so the final answer I get is 9.67 is this correct ? – 2017-02-01
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    @Dan That is what I got :-) – 2017-02-01
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I cannot grasp the actual purpose of such kind of exercises, the idea of the trapezoid method is simple to understand, performing it by hand is just tedious, very error-prone and quite useless, in my personal opinion. Anyway, over the interval $[4,5]$ the function $f(x)=\sqrt{x^3+2}$ behaves like $x^{3/2}$, hence by the Cauchy-Schwarz inequality $$ \int_{4}^{5}x^{3/4}\sqrt{\frac{x^3+2}{x^{3/2}}}\,dx \approx\sqrt{\int_{4}^{5}x^{3/2}\,dx \int_{4}^{5}\frac{x^3+2}{x^{3/2}}\,dx} = \frac{2}{5} \sqrt{3739-1411 \sqrt{5}}$$ and the RHS is about $9+\frac{2}{3}$. That suggests there is an off-by-one in your computations.
The trapezoid method should give you $$ \int_{4}^{5}x^{3/4}\sqrt{\frac{x^3+2}{x^{3/2}}}\,dx\approx\frac{\frac{1}{2}f\left(\frac{16}{4}\right)+f\left(\frac{17}{4}\right)+f\left(\frac{18}{4}\right)+f\left(\frac{19}{4}\right)+\frac{1}{2}f\left(\frac{20}{4}\right)}{4}$$ that is close to $9+\frac{2}{3}$ as well.

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    Many elementary calculus courses just give questions like this to illustrate the behavior of certain quadrature rules. I just don't agree with the choice of function here. – 2017-02-01