Problem

Find the general indefinite integral. $ \disp…

01:43

Problem 8 Easy Difficulty

Find the general indefinite integral.

$ \displaystyle \int (u^6 - 2u^5 - u^3 + \frac{2}{7}) \, du $

Answer

$\frac{u^{7}}{7}-\frac{u^{6}}{3}-\frac{u^{4}}{4}+\frac{2}{7} u+C$

More Answers

00:45

Frank L.

00:36

Amrita B.

Topics

Integrals

Integration

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Discussion

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Video Transcript

the theme in this problem is the integral of some power Is equal to adding 1 to the exponents. And then I like to instead of dividing by the new experiment. Multiply by the reciprocal and then don't forget about plus C. And the whole thought process is the derivative of this thing that I just did is equal to that and the derivative of a constant is zero. So that's why we need a plus C. So as we're looking at this problem of the integral of you to the sips -2 U. to the 5th uh minus U cubed plus two sevens. Do you you're correct answer is going to be just doing each piece individually. Or you add one to the exponent. Multiply by the reciprocal, add one to the exponent. Now instead of believing this is 26 I can reduce to 6 to 1 third, Add 1 to the excrement. Multiply by the reciprocal And add one to the exponent. Well it just becomes like you to the first because again, the director of 272 needs to be this, and like I said earlier, you need to remember plus E. This is your correct answer.

Gregory H.

Topics

Integrals

Integration

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Top Calculus 1 / AB Educators

Catherine R.

Missouri State University

Caleb E.

Baylor University

Samuel H.

University of Nottingham

Joseph L.

Boston College