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Consider the area bounded by $f(x)=x^{2}+1$ and the $x$ -axis, between $x=0$ and $x=b .$ Find $b$ if this area is $1 / 2$.

$$\pm 0.466221$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Oregon State University

Harvey Mudd College

Idaho State University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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So now they tell you the area and this problem is one half. They tell you that the area between the curve and the X axis that's your clue, that you're going to do the integral of its X squared plus one. And what we're doing is they also tell you the balance and the problem are from X equals 02 X equals B. So it makes sense that you do from zero to be for your boundaries. And D X is going to equal one half because this is your equation for the area. And they tell you the area now, before we messing with that one half, it actually makes sense to do the anti director first adding volunteer exponent multiplied rather simple that experiments. Uh, and I'm not going to mess with that one half yet to help us with that shortly because now I can plug in my balance one third be cube plus B uh, and then you can also plug in your lower bounds the zero. But what's nice about plugging in zero is you just get a couple of zeros in there, so that's sort of points. And now I'll examine that equaling one half, but I don't believe this is factory able. It is. I'm sorry, but what I would do is go to a graphing calculator and, uh, you know, plug into one side of the equation. This is one side of the equation, except most graphic calculators don't use like I don't like you used to be, So replace, be with X. And then the second equation. Why? To set equal to one half, which is just a horizontal line. And between those two things, you'll find the point of intersection and whatever the X value is will be your answer for the B as extras replacing B. And it should cross twice the positive and a negative value. Um, I looked around at three decimals. Positive or negative 0.466

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