How do you find the area using the trapezoidal approximation method, given #sinpi*x dx#, on the interval [2, 5] with n=25?

1 Answer
Feb 26, 2017

Answer:

Trapezium rule gives:

# int_2^5 \ sin(pix) \ dx ~~ 0.629062#

Explanation:

The values of #f(x)=sin(pix)# are tabulated as follows (using Excel) working to 5dp

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Using the trapezoidal rule:

# int_a^bydx ~~ h/2{(y_0+y_n)+2(y_1+y_2+...+y_(n-1))}#

We have:

# int_2^5 \ sin(pix) \ dx ~~ 0.119999/2 { (0 - 0.000001) + 2(0.368124#
# " "+ 0.684547 + 0.904827 + 0.998026 + 0.951056 + 0.770513#
# " "+ 0.481753 + 0.125333 - 0.24869 - 0.587786 - 0.844328 #
# " "- 0.982288 - 0.982288 - 0.844328 - 0.587786 - 0.24869#
# " " + 0.125333 + 0.481753 + 0.770513 + 0.951056 + 0.998026 #
# " "+ 0.904827 + 0.684547 + 0.368124) }#

# " "=0.059999 { - 0.000001 + 2(5.242183) }#
# " "=0.059999 { - 0.000001 + 10.484367 }#
# " "=0.059999 { + 10.484367 }#
# " "=0.629062#

Let's compare this to the exact value:

# int_2^5 \ sin(pix) \ dx = [-cos(pix)/pi]_2^5 #
# " " = (-1/pi)[cos(pix)]_2^5 #
# " " = (-1/pi)(cos(5pi)-cos(2pi)) #
# " " = (-1/pi)(-1-1) #
# " " = 2/pi #
# " " ~~ 0.6366197724 ... #