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We arrived at Formula 6.3.2, $ V = \displaystyle …

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Problem 72 Hard Difficulty

(a) Use integration by parts to show that
$$ \int f(x) dx = xf (x) - \int xf^\prime (x) dx $$
(b) If $ f $ and $ g $ are inverse functions and $ f^\prime $ is continuous, prove that
$$ \int_a^b f(x) dx = bf (b) - af (a) - \int_{f(a)}^{f(b)} g(y) dy $$
[Hint: Use part (a) and make the substitution $ y = f(x) $.]
(c) In the case where $ f $ and $ g $ are positive functions and $ b > a > 0 $, draw a diagram to give a geometric interpretation of part (b).
(d) Use part (b) to evaluate $ \displaystyle \int_1^e \ln x dx $.

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Chapter 7

Techniques of Integration

Section 1

Integration by Parts

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Video Transcript

For a the problem is use integration by parts to show that integral of x dx is equal to x, f of x, minus, integral x times, f from x d x. For this problem to select what? U is equal to f of x is equal to 1. Then up is equal to half prix and v is equal to x, then use integration by parts we have the integral of x. Dx is equal to f of x times x, minus integral of 5 f prime x times x dx. So we proved this equation for b. The problem is if f and g are inverse functions and prim is continuous proof that the integral from a to b, f of x dx, is equal to b, f b, minus a f, a minus, the integral f a to f b g y dy. So here we have a hint is part a and make the substitution y is equal to f of x. So if we use substitution y equal to 4 x, then we have x is equal to g of y s. D x is equal to g prime y d y. Now, the integral a to b, f of x dx, is equal to integral, from f of a to f, o b and y g y plus dyand. Then we use the result in part a- and we have this. The integral of this 1 is equal to y g y, from f a to f, b minus, integral from f a to f b to y. We can check what, if both 1 is equal to f, b, j, f b minus f, a times g f, a thinks, f and g are inverse functions to g f b is so. This is f b times b and g f e is a so. This is a times f of a so i proved part b. We see in the case where f and g are positive functions and b is greater or a is greater than 0 jar or aga to give a geometric interpretation of part b. So we can draw a graph here if this is a function as a wax and from a to b by half. This is f over and f of the integral of flax from a to b. Is this part area of this part? This part, this is integral from a to b, f of f, x, dx and integral from f a f b g y d y is its area of this part now b times f b is the rectangle this rectangle and at times f, is this rectangle? So we can see big rectangle, minus the small rectangle and minus area of this part is just the area of this part is just integral from a to b, f x x. This is c use part b to evaluate its integral from 1 to e l, n x dx. So we know y equal to l, n x, inverse function of l n x is e to x, so this is x is equal to e to the integral. From 1 to e n x. Dx is equal to e times. L n minus 1 times on 1 and minus 1 is equal to 0, and e is equal to 1 y is e to dytisequal. To e minus 0 per minus integral to y is tay from 0 to 1. Pooshe answer is 1.

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Problem

We arrived at Formula 6.3.2, $ V = \displaystyle …

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Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 7

Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Joseph Lentino

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Joseph Lentino

James Stewart

Calculus: Early Transcendentals