# Midpoint Rule Calculator

Online calculator for approximating definite integral using the Midpoint (Mid ordinate) Rule, with steps shown.

• In general, you can skip the multiplication sign, so 5x is equivalent to 5*x.
• In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x).
• Also, be careful when you write fractions: 1/x^2 ln(x) is 1/x^2 ln(x), and 1/(x^2 ln(x)) is 1/(x^2 ln(x)).
• If you skip parentheses or a multiplication sign, type at least a whitespace, i.e. write sin x (or even better sin(x)) instead of sinx.
• Sometimes I see expressions like tan^2xsec^3x: this will be parsed as tan^(2*3)(x sec(x)). To get tan^2(x)sec^3(x), use parentheses: tan^2(x)sec^3(x).
• Similarly, tanxsec^3x will be parsed as tan(xsec^3(x)). To get tan(x)sec^3(x), use parentheses: tan(x)sec^3(x).
• From the table below, you can notice that sech is not supported, but you can still enter it using the identity sech(x)=1/cosh(x).
• If you get an error, double check your expression, add parentheses and multiplication signs where needed, and consult the table below.
• All suggestions and improvements are welcome. Please leave them in comments.
The following table contains the supported operations and functions:
 Type Get Constants e e pi pi i i (imaginary unit) Operations a+b a+b a-b a-b a*b a*b a^b, a**b a^b sqrt(x), x^(1/2) sqrt(x) cbrt(x), x^(1/3) root(3)(x) root(x,n), x^(1/n) root(n)(x) x^(a/b) x^(a/b) abs(x) |x| Functions e^x e^x ln(x), log(x) ln(x) ln(x)/ln(a) log_a(x) Trigonometric Functions sin(x) sin(x) cos(x) cos(x) tan(x) tan(x), tg(x) cot(x) cot(x), ctg(x) sec(x) sec(x) csc(x) csc(x), cosec(x) Inverse Trigonometric Functions asin(x), arcsin(x), sin^-1(x) asin(x) acos(x), arccos(x), cos^-1(x) acos(x) atan(x), arctan(x), tan^-1(x) atan(x) acot(x), arccot(x), cot^-1(x) acot(x) asec(x), arcsec(x), sec^-1(x) asec(x) acsc(x), arccsc(x), csc^-1(x) acsc(x) Hyperbolic Functions sinh(x) sinh(x) cosh(x) cosh(x) tanh(x) tanh(x) coth(x) coth(x) 1/cosh(x) sech(x) 1/sinh(x) csch(x) Inverse Hyperbolic Functions asinh(x), arcsinh(x), sinh^-1(x) asinh(x) acosh(x), arccosh(x), cosh^-1(x) acosh(x) atanh(x), arctanh(x), tanh^-1(x) atanh(x) acoth(x), arccoth(x), cot^-1(x) acoth(x) acosh(1/x) asech(x) asinh(1/x) acsch(x)

Enter a function: f=

Enter a lower limit: a=

Enter an upper limit: b=

Enter the number of rectangles: n=

Write all suggestions in comments below.

## Solution

Your input: approximate the integral $$\int_{1}^{3}\sqrt{\sin^{4}{\left (x \right )} + 7}\ dx$$$using $$n=4$$$ rectangles.

The Midpoint Sum (also Midpoint Approximation) uses midpoints of subinterval: $$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)$$$, where $$\Delta{x}=\frac{b-a}{n}$$$.

We have that $$a=1$$$, $$b=3$$$, $$n=4$$$. Therefore, $$\Delta{x}=\frac{3-1}{4}=\frac{1}{2}$$$.

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

Now, we just evaluate the function at those endpoints:

$$f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(1\right)+\left(\frac{3}{2}\right)}{2}\right)=f\left(\frac{5}{4}\right)=\sqrt{\sin^{4}{\left (\frac{5}{4} \right )} + 7}=2.79482192294185$$$$$f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{3}{2}\right)+\left(2\right)}{2}\right)=f\left(\frac{7}{4}\right)=\sqrt{\sin^{4}{\left (\frac{7}{4} \right )} + 7}=2.81735090562718$$$

$$f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(2\right)+\left(\frac{5}{2}\right)}{2}\right)=f\left(\frac{9}{4}\right)=\sqrt{\sin^{4}{\left (\frac{9}{4} \right )} + 7}=2.71413091375118$$$$$f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{5}{2}\right)+\left(3\right)}{2}\right)=f\left(\frac{11}{4}\right)=\sqrt{\sin^{4}{\left (\frac{11}{4} \right )} + 7}=2.64975816351283$$$

Finally, just sum up the above values and multiply by $$\Delta{x}=\frac{1}{2}$$$: $$\frac{1}{2}(2.79482192294185+2.81735090562718+2.71413091375118+2.64975816351283)=5.48803095291652$$$