Question

How do you use the trapezoidal rule and three subintervals to give an estimate for the area between $y=\cos ecx$ and the x axis from the interval $x=\dfrac{\pi }{8}$ to $x=\dfrac{7\pi }{8}$ .

Answer 1

Hint: Now to find the area by trapezoidal rule we will first divide the interval in 3 equal parts. Now we know that the area under the graph is nothing but the addition of the area of each interval. Now to find area of interval [a, b], we will use the formula $\dfrac{f\left( b \right)-f\left( a \right)}{2}\times \left( b-a \right)$ . and hence find the required area.

Complete step by step answer:
Now we want to find the area under the curve by using the trapezoidal method.
Now consider the given integral $\left[ \dfrac{\pi }{8},\dfrac{7\pi }{8} \right]$
Now we will divide the interval into 3 parts such that the width of each interval is $\dfrac{\pi }{4}$
Hence the three intervals are ${{I}_{1}}=\left[ \dfrac{\pi }{8},\dfrac{3\pi }{8} \right]$ , ${{I}_{2}}=\left[ \dfrac{3\pi }{8},\dfrac{5\pi }{8} \right]$ and ${{I}_{3}}=\left[ \dfrac{5\pi }{8},\dfrac{7\pi }{8} \right]$ .
Now let us first find the value of the function at each edge of the interval with the help of a trigonometric sheet.
$\begin{align}
  & \cos ec\left( \dfrac{\pi }{8} \right)=2.613125 \\
 & \cos ec\left( \dfrac{3\pi }{8} \right)=1.082392 \\
 & \cos ec\left( \dfrac{5\pi }{8} \right)=1.0823922 \\
 & \cos ec\left( \dfrac{7\pi }{8} \right)=2.613125 \\
\end{align}$
Now the area of each interval [a, b] is calculated as $\dfrac{f\left( b \right)+f\left( a \right)}{2}\times \left( b-a \right)$ where $\left( b-a \right)$ is the width of the interval.
Hence we get area of intervals as,
Now area of ${{I}_{1}}=\dfrac{2.613125+1.082392}{2}\times \dfrac{\pi }{4}=0.4619\pi $
Area of interval ${{I}_{2}}=\dfrac{1.082392+1.082392}{2}\times \dfrac{\pi }{4}=0.2705\pi $
And Area of interval ${{I}_{3}}=\dfrac{1.802392+2.613125}{2}\times \dfrac{\pi }{4}=0.5519\pi $
Now area under the curve is given by ${{I}_{1}}+{{I}_{2}}+{{I}_{3}}$
Hence area under the curve is $0.4619\pi +0.2705\pi +0.5519\pi =1.2843\pi $
Hence the area under the curve is $1.2843\pi $ .

Note:
Note that this method gives us just the approximate value of integral. Here we consider the interval to be a rectangle and hence we divide the graph into several rectangles. Now we know that the area of the rectangle is length × height. Here length is nothing but width of the graph and we take height as average value of the function in the by considering the values of the function at the endpoints of the interval.