Calculate The Iterated Integral T 2sin 3 S

The iterated integral of t 2sin(3s) is a fundamental concept in calculus that involves integrating a function with respect to one variable and then integrating the result with respect to another variable. This process is essential in solving problems in physics, engineering, and other scientific disciplines where functions depend on multiple variables.

What is the iterated integral t 2sin 3 s?

The iterated integral of t 2sin(3s) refers to the process of integrating the function t 2sin(3s) with respect to one variable and then integrating the result with respect to another variable. This is commonly written as:

∫[a to b] ∫[c to d] t 2sin(3s) dt ds

This double integral represents the volume under the surface defined by t 2sin(3s) over the region [a, b] × [c, d] in the ts-plane. The iterated integral can be evaluated using the Fundamental Theorem of Calculus by first integrating with respect to t and then with respect to s.

How to calculate the iterated integral

Calculating the iterated integral of t 2sin(3s) involves the following steps:

First, integrate t 2sin(3s) with respect to t from t = a to t = b.
Then, integrate the result with respect to s from s = c to s = d.

This process is known as iterated integration or repeated integration. The order of integration can sometimes be reversed, but the result will be the same if the limits of integration are adjusted accordingly.

The formula explained

The iterated integral of t 2sin(3s) can be expressed using the following formula:

∫[a to b] ∫[c to d] t 2sin(3s) dt ds = ∫[c to d] [ (t³/3) sin(3s) ] evaluated from t = a to t = b ds

This formula shows that the inner integral with respect to t results in a function of s, which is then integrated with respect to s. The final result is the volume under the surface defined by t 2sin(3s) over the specified region.

Worked example

Let's calculate the iterated integral of t 2sin(3s) from t = 0 to t = 1 and s = 0 to s = π/3.

First, integrate t 2sin(3s) with respect to t from t = 0 to t = 1:
∫[0 to 1] t 2sin(3s) dt = [ (t³/3) sin(3s) ] evaluated from t = 0 to t = 1 = (1³/3) sin(3s) - (0³/3) sin(3s) = (1/3) sin(3s)
Next, integrate the result with respect to s from s = 0 to s = π/3:
∫[0 to π/3] (1/3) sin(3s) ds = (1/3) [ -cos(3s)/3 ] evaluated from s = 0 to s = π/3 = (1/9) [ -cos(π) - (-cos(0)) ] = (1/9) [ -(-1) - (-1) ] = (1/9) [ 1 + 1 ] = 2/9

The final result of the iterated integral is 2/9.

FAQ

What is the difference between a single integral and an iterated integral?: A single integral calculates the area under a curve, while an iterated integral calculates the volume under a surface over a region in the plane.
Can the order of integration be reversed in an iterated integral?: Yes, the order of integration can sometimes be reversed, but the limits of integration must be adjusted accordingly to maintain the same volume.
What are some practical applications of iterated integrals?: Iterated integrals are used in physics to calculate work done by variable forces, in engineering to find the center of mass of a variable-density object, and in probability to calculate expected values.