Finding area Find the area of the region enclosed by the...

Finding area Find the area of the region enclosed by the curve $y=x \sin x$ and the $x$ -axis (see the accompanying figure) for a. $0 \leq x \leq \pi$ b. $\pi \leq x \leq 2 \pi$ c. $2 \pi=r=3 \pi$ d. What pattern do you see here? What is the area between the curve and the $x$ -axis for $n \pi \leq x \leq(n+1) \pi, n$ an arbitrary nonnegative integer? Give reasons for your answer. (FIGURE CAN'T COPY)

Finding area Find the area of the region enclosed by the curve $y=x \sin x$ and the $x$ -axis (see the accompanying figure) for
a. $0 \leq x \leq \pi$
b. $\pi \leq x \leq 2 \pi$
c. $2 \pi=r=3 \pi$
d. What pattern do you see here? What is the area between the curve and the $x$ -axis for $n \pi \leq x \leq(n+1) \pi, n$ an arbitrary nonnegative integer? Give reasons for your answer.
(FIGURE CAN'T COPY)

Key Concepts

Integration by Parts

Integration by parts is a technique used to integrate products of functions, based on the product rule for differentiation. It is particularly useful when one part of the product simplifies upon differentiation while the other is easily integrable. This method is often employed when handling integrals of functions that are the product of an algebraic term and a trigonometric function.

Handling Absolute Values in Integration

When integrating a function that changes sign over an interval, it is necessary to account for the absolute value of the function to ensure that all computed areas are positive. This involves dividing the interval into subintervals where the function maintains a consistent sign and integrating the absolute value over each subinterval, then summing the results.

Periodicity and Symmetry in Functions

Many trigonometric functions exhibit periodic behavior and symmetry, which can simplify the analysis of areas over repeated intervals. Recognizing and exploiting patterns in the integral or area as the interval moves through periods can lead to general formulas. This pattern recognition is useful not only in solving specific problems but also in understanding the underlying regular behavior of periodic functions.

Area Between a Curve and the x-axis

Finding the area between a curve and the x-axis often requires careful attention to the sign of the function. When a function dips below the x-axis, its integral is negative; however, the area is always a nonnegative quantity. To accurately measure the total area, one must either integrate the absolute value of the function or break the interval into segments over which the function does not change sign.

Definite Integrals

A definite integral calculates the accumulation of a quantity, such as area, between two limits. It is the fundamental tool for finding the net signed area under a curve by integrating a function over an interval. This concept is crucial when computing areas bounded by curves and the x?axis, as it involves setting up and evaluating an integral with proper limits.

Transcript

00:01 To find the areas between the reasons which are indicated in the question.

00:14 So we have the party of the question which states integration of 0 to pi x, sine x d x.

00:27 So it is given as x integration of sine x d x minus d upon d x of x of x x of x.

00:37 Integration of sinexx d x or d x.

00:43 This is given as x into minus cos x minus equation of y into minus plus x this is given as minus x cos x plusx plus x plus x plus plus x plus x plus x plus x plus since x and intervals are 0 to pi so after putting the values of 0 and pi we have this as minus pi minus 1 plus sine pi minus 0 plus sine by minus 0 this is given as pi plus 0 this given as pi so we have the answer for part of the problem now part of the problem stills the integration of x, sine x d x, but the curve is towards the lower half of the x -ex is therefore we have to put minus sign over here and the range is pi to two to pi so the integration is minus of minus x, cos x, minus sine x you have to put the limits as pi and 2 pi so it is given as x x x minus sine x pi and 2 pi so so putting the values we have this as 2 pi now this is given as 2 pi into 1 minus 0 minus 5 into minus 5 minus 1 minus 1 plus 0 so it is 2 pi plus pi this is given as 3 pi so it is the answer for be part b of the problem now we have part c of the question which states the division of the division of x sine x t x and the radius 2 pi to 3 pi so it is given as minus x cos x plus the range is 2 pi to 3 pi so this is given as minus 3 pi cos 3 pi plus sine 3 pi minus of minus 2 pi close 2 pi plus sine 2 pi plus sine 2 pi this is given as minus 3 pi and cos 3 pie equals minus 1 plus sine 3 is 0 minus of minus 2 pi and cost 2 pi is 1 plus sine 2 pi is 0 this is given as 3 pi plus 2 pi which is equal to 5 pi so this is the answer to the part see of the problem now we have the last part which is a addition of x sine x t x the range is n pi 2 n plus 1 so answer is minus of minus x, cos x plus sine x.

05:14 The range is n this is n pi and is n plus 1.

05:25 It is given as x x x x minus sin x the range is n pi to n plus 1 pi.

05:42 This is given as n plus 1.

05:45 5, cost of n plus 1, 5 minus sign of n plus 1, pi minus n plus 1, pi minus n pi, n pi, cos n5, minus sign, n pi...

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