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Evaluate the given integral and check your answer.$$\int \frac{1}{2 w^{3}} d w$$

$$-\frac{1}{4 w^{2}}+c$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 1

Anti differentiation - Integration

Integrals

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

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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Yeah. Okay, So this is a good problem because it gets students toe, recognize the mistakes with rational exponents. So we look at one or two W cute D w Andi. I always have my students to rewrite this with a rational exponents and because w is in the denominators to the native third power. Now, this is where students get it wrong. Is that, too is still in the denominator. Because remember, the order of operations says to do the exponents before multiplying. So that means Onley w is taken to that power D w So when we do the anti derivative and we add one to our experiment just a reminder. Negative three plus one is negative two. So I knew experts negative. Two way still have that one half from the problem. But then I need to multiply by the reciprocal of negative too. So negative one half. Uh, I'm multiplying in there on Don't forget about plus C now, I would clean this up. Just a me habit. Your teacher might let you legally and so like that. But if you have a positive times negative a negative 1/4 and I would stick that w back into the denominator with that negative extra. Plus C. Now, if I were checking my work, I would go back to the derivative of this one and bring that exponents front, which would cancel out with that negative too. So I would be back at one half w to the native third power in the derivative of a constant zero. So my work here in blue is checking my work. Which brings me back to this. Which brings me back to this. Which means I did this correctly. And my answer in grain was correct. Yeah.

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