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# Trapezoidal Rule Calculator

Calculator will approximate integral using trapezoidal rule.

Enter a function: $f=$

Enter lower limit: $a=$

Enter upper limit: $b=$

Enter number of rectangles: $n=$

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Solution

Your input: approximate integral $\displaystyle{\int_{0}^{1}\sqrt{\sin^{3}{\left (x \right )} + 1}\ dx}$ using $n=5$ rectangles.

Trapezoidal rule states that $\displaystyle{\int_{a}^{b}f(x)dx}\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)$, where $\displaystyle{\Delta{x}=\frac{b-a}{n}}$.

We have that $a=0$, $b=1$, $n=5$.

Therefore, $\displaystyle{\Delta{x}=\frac{1-0}{5}}=\frac{1}{5}$.

Divide interval $\left[0,1\right]$ into $n=5$ subintervals of length $\Delta{x}=\frac{1}{5}$ with the following endpoints: $a=0, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, 1=b$.

Now, we just evaluate function at those endpoints:

$f(x_{0})=f(a)=f(0)=1=1$

$2f(x_{1})=2f(\frac{1}{5})=2 \sqrt{\sin^{3}{\left (\frac{1}{5} \right )} + 1}=2.007826067912793$

$2f(x_{2})=2f(\frac{2}{5})=2 \sqrt{\sin^{3}{\left (\frac{2}{5} \right )} + 1}=2.058206972332648$

$2f(x_{3})=2f(\frac{3}{5})=2 \sqrt{\sin^{3}{\left (\frac{3}{5} \right )} + 1}=2.17257446116512$

$2f(x_{4})=2f(\frac{4}{5})=2 \sqrt{\sin^{3}{\left (\frac{4}{5} \right )} + 1}=2.340214753424868$

$f(x_{5})=f(b)=f(1)=\sqrt{\sin^{3}{\left (1 \right )} + 1}=1.263258974474734$

Finally, just sum up above values and multiply by $\displaystyle\frac{\Delta{x}}{2}=\frac{1}{10}$: $\frac{1}{10}(1+2.007826067912793+2.058206972332648+2.17257446116512+2.340214753424868+1.263258974474734)=1.084208122931016$

Answer: $1.084208122931016$.

Related Note: Trapezoidal Rule