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(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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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

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Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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(a) Use integration by par…

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a. Use integration by part…

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A useful integrala. Us…

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Prove that $\int_{a}^{b} \…

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Use integration by parts …

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Use integration by parts t…

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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