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Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves.

$ y = x \ln (x + 1) $ , $ y = 3x - x^2 $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Missouri State University

Campbell University

Oregon State University

Harvey Mudd College

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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Use a graph to find approx…

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The problem is use a graph to find approximate x coordinates of the points of intersection with given curves then find the area of the region bondage curves here y is equal to x times. L n, x, plus 1 and y is equal to 3 x minus x. Squared by using some graphic device we can find, the graph is as follows. So y 1 is y, is equal to x, times, x, plus 1 and a blue 1, as is equal to 3 x minus x square, and we can see its intersection. Points are 0 and 1.926, so we can let a is equal to 0 and b is equal to 1.926. Then, and we can see between these 2 points. The blue curve is on top of the right 1. So we have every bonds. 2 curves is equal to integral, from a to e 3 x, minus x, squared minus x times n x, plus 1 dx. So this is equal to 3 over 2 x, squared minus 1, half 11 third to 3 power from a to b minus integral from a to b x times, l n x, plus 1 dx. But then we compute the integral of x times x, plus 1. I see we kind o use integration by parts, but first we cannot. Y is equal to x, plus 1 used substitution and we have. This is equal to the integral of y minus 1 times n x, o n y d y, and then we can use? U substitution, that? U is equal to l, n y and prime, is equal to y minus 1. Then u, prime, is equal to 1. Over y and v is equal to 1 half y squared and minus y. Then this integral is equal to u times v. So this is 1 half y squared minus y times n y minus integral of prim times v. So this is 1 half y minus 1. This is equal to 1 half y squared minus y n y minus. This is upon force y square, minus y and as cos, the number c now and we convert y back to x. So this is 1 half x, plus 1 square minus x, plus 1 times l n x, plus 1 minus a force x, plus 1 square minus x, plus 1 and for our problem is integral definitely integral from a to b of this value. So this is from a to b, then this is equal to this 1 and from a to b now, by using a calculator we can find. The value of this 1 is about 1.69 twoix.

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