Answer
$$
\int_{0}^{4}\sqrt{y} \cos y dy ,\quad n=8
$$
Here
$$
f(y)=\sqrt{y} \cos y ,\quad \Delta y=\frac{4-0}{8}=\frac{1}{2}
$$
(a) the Trapezoidal Rule
$$
\begin{aligned}
T_{8} &=\frac{1}{2 \cdot 2}\left[f(0)+2 f\left(\frac{1}{2}\right)+2 f(1)+2 f\left(\frac{3}{2}\right)+2 f(2)+\\
+2 f\left(\frac{5}{2}\right)+2 f(3)+2 f\left(\frac{7}{2}\right)+f(4)\right] \\
&\approx-2.364034
\end{aligned}
$$
(b) The midpoints Rule is given by:
$$
\begin{aligned}
M_{8} &=\frac{1}{2}\left[f\left(\frac{1}{4}\right)+f\left(\frac{3}{4}\right)+f\left(\frac{5}{4}\right)+f\left(\frac{7}{4}\right)+\\
\quad\quad +f\left(\frac{9}{4}\right)+f\left(\frac{11}{4}\right)+f\left(\frac{13}{4}\right)+f\left(\frac{15}{4}\right)\right]\\
& \approx-2.310690
\end{aligned}
$$
(c) Simpson’s Rule:
$$
\begin{aligned}
S_{8} & =\frac{1}{2 \cdot 3}\left[f(0)+4 f\left(\frac{1}{2}\right)+2 f(1)+4 f\left(\frac{3}{2}\right)+\\
\quad\quad +2 f(2)+4 f\left(\frac{5}{2}\right)+2 f(3)+4 f\left(\frac{7}{2}\right)+f(4)\right] \\
& \approx-2.346520
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{4}\sqrt{y} \cos y dy ,\quad n=8
$$
Here
$$
f(y)=\sqrt{y} \cos y ,\quad \Delta y=\frac{4-0}{8}=\frac{1}{2}
$$
and so the Trapezoidal Rule can be given by:
$$
\int_{a}^{b} f(x) d x \approx T_{n}=\frac{\Delta x}{2}\left[f\left(x_{0}\right)+2 f\left(x_{1}\right)+2 f\left(x_{2}\right)+\cdots+2 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right]
$$
Therefore,
(a) the Trapezoidal Rule
$$
\begin{aligned}
T_{8} &=\frac{1}{2 \cdot 2}\left[f(0)+2 f\left(\frac{1}{2}\right)+2 f(1)+2 f\left(\frac{3}{2}\right)+2 f(2)+\\
+2 f\left(\frac{5}{2}\right)+2 f(3)+2 f\left(\frac{7}{2}\right)+f(4)\right] \\
&\approx-2.364034
\end{aligned}
$$
(b) The midpoints Rule is given by:
$$
\int_{a}^{b} f(x) d x \approx M_{n}=\Delta x\left[f\left(\bar{x}_{1}\right)+f\left(\bar{x}_{2}\right)+\cdots+f\left(\bar{x}_{n}\right)\right]
$$
where
$$
\bar{x}_{i}=\frac{1}{2}\left(x_{i-1}+x_{i}\right)=\text { midpoint of }\left[x_{i-1}, x_{i}\right]
$$
So, we have:
$$
\begin{aligned}
M_{8} &=\frac{1}{2}\left[f\left(\frac{1}{4}\right)+f\left(\frac{3}{4}\right)+f\left(\frac{5}{4}\right)+f\left(\frac{7}{4}\right)+\\
\quad\quad +f\left(\frac{9}{4}\right)+f\left(\frac{11}{4}\right)+f\left(\frac{13}{4}\right)+f\left(\frac{15}{4}\right)\right]\\
& \approx-2.310690
\end{aligned}
$$
(c) Simpson’s Rule:
$$
\begin{aligned}
\int_{a}^{b} f(x) d x & \approx S_{n} \\
&=\frac{\Delta x}{3}\left[f\left(x_{0}\right)+4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)\\
+\cdots\right. \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right]
\end{aligned}
$$
So, we obtain:
$$
\begin{aligned}
S_{8} & =\frac{1}{2 \cdot 3}\left[f(0)+4 f\left(\frac{1}{2}\right)+2 f(1)+4 f\left(\frac{3}{2}\right)+\\
\quad\quad +2 f(2)+4 f\left(\frac{5}{2}\right)+2 f(3)+4 f\left(\frac{7}{2}\right)+f(4)\right] \\
& \approx-2.346520
\end{aligned}
$$
