Question

integrate cos theta× d theta whose upper limit is 2π and lower limit is π? Please answer this question​

Answer 1

Answer:

Using the definition of the integral and the fact that sinx is an odd function, from 0 to 2π, with equal area under the curve at [0,π] and above the curve at [π,2π], the integral is 0.

This holds true for any time sinx is evaluated with an integral across a domain where it is symmetrically above and below the x-axis.

∫2π0sinxdx=[−cosx]∣∣∣2π0

=−cos2π−(−cos0)

An alternative way to do this starting from the limit definition is:

∫baf(x)dx=limn→∞N∑i=1f(x*i)Δx

where:

n is the number of rectangles used to approximate the integral, i.e. the area between the curve and the x-axis.

i is the index of each rectangle in [0,2π].

N is the index of the final rectangle in [0,2π].

f(x*i) is the height of each given rectangle in [0,2π], which varies as sin(x).

Δx is the width of each given rectangle in [0,2π], which converges to 0 as n→∞.

If we use the midpoint-rectangular approximation method (MRAM), we choose a convenient interval Δx such that we can find a midpoint for each rectangle of dimension Δx×f(x*i), where the midpoint of the ith rectangle is defined as

Mi=xi−1+xi−xi−12.

Let us choose Δx=π2 such that x={0,π2,π,3π