anitderivatives to x+2, square root of (25-x^2), 2-x, and

[chillintoucan28]

please tell me the antiderivatives to the follofing equations and show me you found each antiderivative

x+2, square root of (25-x^2), 2-x, absolute value of (x-2)

 

[galactus]

Re: anitderivatives

The first and third are very basic. The second one can be done with a trig sub.

\(\displaystyle \int\sqrt{25-x^{2}}dx\)

Let \(\displaystyle x=5sin(t), \;\ dx=5cos(t)dt\)

Then we get:

\(\displaystyle 25\int\cos^{2}(t)dt\)

The fourth one, we can use a little trick for this form.

When you have an integral of the form \(\displaystyle \int |x-a|dx\), then we can integate as if there were no absolute value and get:

\(\displaystyle \frac{x^{2}}{2}-ax\)

Then add \(\displaystyle \frac{a^{2}}{2}\) to it:

\(\displaystyle \frac{x^{2}}{2}-ax+\frac{a^{2}}{2}=\frac{(x-a)^{2}}{2}\)

Now, this equals \(\displaystyle \frac{(x-a)(x-a)}{2}\)

Now, make one of them an absolute:

\(\displaystyle \frac{(x-a)|x-a|}{2}\)

In your case x=2.

 

[chillintoucan28]

Re: anitderivatives

could you please do the first and third one for me...

 

[tkhunny]

Re: anitderivatives

Please show your work. No effort or work shown by the student never is a good sign.

Let's see how you solve them.

 

[chillintoucan28]

Re: anitderivatives

ok well... i just need help on the antiderivative of 2-x...

2-x = -x+2

then we use the power rule of antirderivatives


so we get (-x^2)/2 + C

C= +2x

so the antiderivative is (-x^2)/2 + 2x, is that right?

 

[tkhunny]

Re: anitderivatives

1) Why did you switch it around? 2 - x is perfectly acceptable.

\(\displaystyle \int 2\;dx = 2x + C\)

\(\displaystyle \int -x\;dx = -\frac{x^{2}}{2} + C\)

Put them together and what do you get?

Note: 'C' is an arbitrary constant. We often don't care what it is. Just knowing that there is such a thing is the important part.

 

[chillintoucan28]

Re: anitderivatives

sorry, im sort of confused, what do you mean "put them together"?

Im trying to find the defintie integral from x=0 to x=4 for the equation f(x) = 2-x, that is qhy I am trying to find the antiderivative, so I can use the fundamental theorem of calculus

 

[stapel]

Im trying to find the defintie integral from x=0 to x=4 for...2-x....

Apply the Power Rule: For x[sup:2p1wl2ed]n[/sup:2p1wl2ed], the antiderivative is (x[sup:2p1wl2ed]n+1[/sup:2p1wl2ed]) /(n + 1). :wink:

Eliz.

 

[Deleted member 4993]

\(\displaystyle \int (2-x) dx \, = \, \frac{-(2-x)^2}{2} + C\)

Evaluate the function above at x = 4

Evaluate the function above at x = 0

Subtract

That's it.....

 

[chillintoucan28]

how did you get that equation, did you use the power rule, could you please show work to how you got it, it would be greatly appreciated

 

[tkhunny]

You should have command of simple algebra. "Putting together" has other names, most notably "addition".

\(\displaystyle \int 2-x\;dx\;=\;\int 2\;dx\;-\;\int x\;dx\)

Do them together or do them separately.

 

[chillintoucan28]

why is there a (-) sign in fron of (2-x)^2/2, isn' the antiderivative simply (2-x)^2/2, why is there the need for a negative sign and where did it come from?

 

[Deleted member 4993]

why is there a (-) sign in fron of (2-x)^2/2, isn' the antiderivative simply (2-x)^2/2, why is there the need for a negative sign and where did it come from?

What is the antiderivative of '-x'?

 

[chillintoucan28]

-x^2/2, but why does that matter?

 

[Deleted member 4993]

-x^2/2, but why does that matter?

That is correct - but tell me why do you have a negative sign in front of "x^2/2"?

Have you done integration by substitution?

If you have then try substituting

u = 2 - x

du = ???

Complete...

 

[chillintoucan28]

i have not done integration by substitution, look i just need the antiderivative of 2-x, is it or is it not (-x^2)/2 + 2x, if it is not please could you show me the correct answer and how you got it?

I got my answer by using the power rule of antiderivatives

look, school has not started yet, I am doing math for my personal benefit, you do not have to be so firm on having me solve the problem on my own...

 

[Deleted member 4993]

i have not done integration by substitution, look i just need the antiderivative of 2-x, is it or is it not (-x^2)/2 + 2x, if it is not please could you show me the correct answer and how you got it?

I got my answer by using the power rule of antiderivatives

look, school has not started yet, I am doing math for my personal benefit, you do not have to be so firm on having me solve the problem on my own...

I only insist that you understand what you are doing.

-(2-x)^2/2 = (-x^2 + 4x - 4)/2 = -x^2/2 + 2x + 1

It is the same answer you got for anti-derivative - differs by a constant.

 

[chillintoucan28]

could you please throughly show me how to do that?

 

[stapel]

could you please throughly show me how to do that?

I'm sorry, but we simply cannot teach courses here, even if the student has the necessary background (in this case, a good grasp of differential calculus). If you are attempting self-study, that's great, but to obtain actual course instruction, you'll need to enroll in a course or else hire a qualified local tutor with whom you can meet for private lessons.

My best wishes to you in your studies!

Eliz.

 

[chillintoucan28]

ok foreget about 2-x, can some be kind enough to show me how to find the antiderivative of square root of (25-x^2), thats my very last question, I promise, but I do need a through step by step process shown, Thank You!