Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

Why is integral of y = x^2 + b , 1/3 x^3 + b but integral of y = x + b is 1/2 x^2 + x ? sorry if my question is simple or incorrect :) i'm new at calculus :)

OpenStudy (australopithecus):

is b a constant?

OpenStudy (anonymous):

yes :)

terenzreignz (terenzreignz):

Hey... uhh \[\Large \int (x^2+b)dx = \frac13x^3 + bx \color{orange}{+C}\]

terenzreignz (terenzreignz):

Wrong emphasis.. I should have emphasised the bx part.

OpenStudy (anonymous):

sorry i don't understand :)

terenzreignz (terenzreignz):

Well the integral of a constant is just THAT constant times x. So the integral of the b part is bx, not simply b.

OpenStudy (australopithecus):

take the derivative of your answer if it doesnt give you what you integrated then you did it wrong

terenzreignz (terenzreignz):

Actually^ LOL

OpenStudy (australopithecus):

Im so rusty with integrals blah I need to relearn them they are a lot of fun

OpenStudy (anonymous):

so my question is wrong?

OpenStudy (australopithecus):

what is the derivative of 1/3 x^3 + b

OpenStudy (australopithecus):

also you didnt include + c so you were wrong even if your answer was right

OpenStudy (anonymous):

i think it's x+b ?

OpenStudy (anonymous):

perhaps i made mistake

OpenStudy (australopithecus):

someone is going to have to go over this with him I'm too rusty and have a journal to read on some journal article on some histone demethylation protein that is in rice that I'm suppose to care about

terenzreignz (terenzreignz):

Okay... let's recap: \[\Large \int(x^2+b)dx\]

terenzreignz (terenzreignz):

You know that antidifferentiation (integration) is linear, so we can distribute it: \[\Large = \int x^2dx + \int b dx\] Right?

terenzreignz (terenzreignz):

Now all you need to recall is the power rule for integration... \[\Large \int ax^ndx = \frac{ax^{n+1}}{n+1}\color{orange}{+C} \qquad n \ne -1\]

OpenStudy (anonymous):

no, i think i should learn more

terenzreignz (terenzreignz):

Not forgetting of course, that \[\Large b = bx^0\]

OpenStudy (anonymous):

sorry wait for about...some days i think i have to learn more,and ask some questions :)

OpenStudy (anonymous):

i'll keep this question open

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!