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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 = \arcsin \left(\frac{1}{2} x \right) $, $ y = 2 - x^2 $

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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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Use a graph to find approx…

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The problem is use a graph to find approx. Approximate x, coordinates of the points of intersection of given curves then find it area of the region bounded by the curves curve is y, is equal to sine x. Over 2 and y is equal to 2 minus x square. The first lesson: look at the graph of these 2 curves. The right 1 is y is equal to oxide x over 2 and the blue 1 is equal to 2 minus x square. We can see its intersection point is negative 1.7551 and 1.1 sen 2, and between these 2 points the blue curve is greater than the red curve. So we have a is equal to negative 1.751 and b is equal to 1.12, and i ride area of the region. Bonded by the curves, as integral from a to b 2 minus x, squared minus oxine, half x dx. So this is equal to so. The integral of 2 minus x squared is equal to 2 x, minus 1 third x to 3 power from a to b minus the integral from a to b oxide 1 half x x. For this integral. We can use integration by parts that, u is equal to oxide x over 2 and v. Prim is equal to 1. It then is equal to 1 over rotie 1 minus x, squared over 4 times by half. So this is equal to 1 over root of 4 minus x, squared and v is equal to x. So this integral is equal to x times oxide 1, over 2 times x, from a to b minus integral x, over relative 4 minus x, squared dx, and for this integral we can use. U substitution, not y is equal to 4 minus x. Squared and y is equal to negative 2 x dx, so we can write this integral is equal to negative, integral from a to b 1 half du over relative. U so this is a to negative root of? U so. This is root of 4 minus x square. From a to b, then we can write. Area of this region is equal to area of this region is 2 minus x square minus sine 1 over 2 x x, so this is equal to 2 x, minus 1 third x 2 cube minus oxide x over 2 times x and plus retto 4 minus x Square from a to b by using calculator we can find this is equal to 3.9993.

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