/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Integration by Substitution Use ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Integration by Substitution Use integration by substitution to show that if \(y\) is a continuous function of \(x\) on the interval \(a \leq x \leq b,\) where \(x=f(t)\) and \(y=g(t),\) then $$\begin{array}{l}{\int_{a}^{b} y d x=\int_{t_{1}}^{t_{2}} g(t) f^{\prime}(t) d t} \\ {\text { where } f\left(t_{1}\right)=a, f\left(t_{2}\right)=b, \text { and both } g \text { and } f^{\prime} \text { are continuous }} \\ {\text { on }\left[t_{1}, t_{2}\right].}\end{array}$$

Short Answer

Expert verified
The transformed integral under the substitution is \( \int_{t1}^{t2} g(t) f'(t) dt \), which holds true if both 'g' and \( f'(t) \) are continuous on \( [t1, t2] \).

Step by step solution

01

Transformation to Variables in Terms of 't'

Let \( y = g(t) \) and \( x = f(t) \). Now substitute these in the original integral. We currently have \( \int_{a}^{b} y dx \) which will transform to \( \int_{a}^{b} g(t) dx(t) \). In the new integral let's replace \( dx \) using the derivative \( dx = f'(t) dt \). After substitution we have \( \int_{a}^{b} g(t) f'(t) dt \). Note that as 'x' changes from 'a' to 'b', 't' changes from 't1' to 't2'.
02

Change of Limits

For integral with respect to 't', we need to change the limits of the integral using the conditions given in the problem. When \( x=f(t) \), if \( x=a, then f(t)=a implies t=t1 \). Similarly when \( x=b, we get t=t2 \). So final integral with respect to 't' becomes \( \int_{t1}^{t2} g(t) f'(t) dt \).
03

Verify the Conditions

The function 'g' is given to be continuous on \( [t1, t2] \), as stated in the problem. Also the function \( f'(t) \) has to be continuous in \( [t1, t2] \). 'f' being a differentiable function implies \( f'(t) \) exists and hence is continuous. Thus the final integral is well defined.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Change of Variables
The change of variables technique, also known as integration by substitution, is a powerful tool that allows us to transform an integral into an easier one to compute. Imagine you have a continuous function, like in our exercise, where the variable of integration is more complex. By changing variables, you can simplify this process.
In our exercise, we let \( y = g(t) \) and \( x = f(t) \). This change rephrases the problem into terms of \( t \), which may be more manageable. The original integral \( \int_{a}^{b} y \, dx \) then becomes \( \int_{a}^{b} g(t) \, dx(t) \). By substituting \( dx = f'(t) \, dt \), we are able to express the entire integral in terms of \( t \). Such transformations are essential because they can transform difficult integrals into standard forms that we already know how to solve.
Remember, whenever you perform a change of variables:
  • Ensure that the new variable is suitably defined over the same interval.
  • Adapt the limits of integration accordingly.
This method is especially useful when dealing with integrals involving composite functions or when integrals are not straightforward to evaluate in their initial form.
Continuous Functions
Continuous functions are integral to using substitution effectively. A function is continuous if small changes in the input of the function lead to small changes in the output. This property assures that no sudden "jumps" or "gaps" occur in the function over its domain.
In the context of our exercise, both \( g(t) \) and the derivative \( f'(t) \) need to be continuous over the interval \([t_1, t_2]\). This requirement is important because when performing integration by substitution, we derive new functions, and we rely on their continuity for the transformations to be valid.
Continuous functions ensure the integral remains well-defined. Without continuity, the integral may encounter points of discontinuity, which can complicate or sometimes invalidate the result. Thus, always check the continuity of your functions before advancing with the substitution process. Here are key elements to consider about continuous functions:
  • They are predictable, without unexpected changes in value.
  • They ensure the smoothness of the resultant integral calculation.
Definite Integral
A definite integral, such as \( \int_{a}^{b} y \, dx \), is a fundamental concept in calculus that represents the accumulation of quantities, like areas under curves, between two bounds \( a \) and \( b \). In our exercise, this is transformed through substitution into \( \int_{t_1}^{t_2} g(t) f'(t) \, dt \).
Definite integrals have precise limits, indicating the start and end points of the integration process. This sets them apart from indefinite integrals, which are concerned more with general antiderivatives.
In integration by substitution, adjusting the limits from \([a, b]\) to \([t_1, t_2]\) is crucial. This adaptation corresponds to the transformed function and its new range. For example, as \( x \) moves from \( a \) to \( b \), the transformed variable \( t \) shifts accordingly from \( t_1 \) to \( t_2 \) through the function \( f(t) \).
In summary, working with definite integrals:
  • Ensure that the limits match the variable of integration utilized.
  • Interpret the results in the context of the problem, such as areas under curves or total accumulated change.
This method lets us calculate integrated values accurately for specific intervals, benefiting from knowing the precise boundary conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Finding a Polar Equation In Exercises \(33-38\) , find a polar equation for the conic with its focus at the pole and the given eccentricity and directrix. (For convenience, the equation for the directrix is given in rectangular form.) $$\begin{array}{ll}{\text { Conic }} & {\text { Eccentricity }} \\ {\text { Parabola }} & {e=1}\end{array} \begin{array}{l}{\text { Directrix }} \\\ {x=-3}\end{array}$$

Finding the Area of a Polar Region Between Two Curves In Exercises \(37-44\) , use a graphing utility to graph the polar equations. Find the area of the given region analytically. Common interior of \(r=2(1+\cos \theta)\) and \(r=2(1-\cos \theta)\)

Finding an Angle In Exercises \(107-112,\) use the result of Exercise 106 to find the angle \(\psi\) between the radial and tangent lines to the graph for the indicated value of \(\theta\) . Use a graphing utility to graph the polar equation, the radial line, and the tangent line for the indicated value of \(\theta .\) Identify the angle \(\psi\) . \(r=2(1-\cos \theta) \quad \theta=\pi\)

Area of a Region In Exercises \(55-58\) , use the integration capabilities of a graphing utility to approximate the area of the region bounded by the graph of the polar equation. $$r=\frac{2}{7-6 \sin \theta}$$

Identifying and Sketching a Conic In Exercises \(13-22\) , find the eccentricity and the distance from the pole to the directrix of the conic. Then identify the conic and sketch its graph. Use a graphing utility to confirm your results. $$r=\frac{24}{25+25 \cos \theta}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.