Simply Checking an Answer in Calculus

[The Bob]

Hello all, again.

Simple question.

Compute \int_{9}^{4} (x - \frac{1}{\sqrt{x}}) dx

I did this:

\int_{9}^{4} (x - \frac{1}{x^{0.5}}) dx = \int_{9}^{4} (x - x^{-0.5}) dx

\left[ \frac{1}{2} x^2 - 2 x^{0.5} \right]_{9}^{4} = \left[ \frac{x^2}{2} - 2 \sqrt{x} \right]_{9}^{4}

(\frac{81}{2} - 6) - (\frac{16}{2} - 4) = \frac{81 - 12}{2} - \frac{16 - 8}{2} = \frac{69}{2} - \frac{8}{2} = \frac{61}{2} = 30.5

2 months of calculus and and still I am not sure of what I have done wrong. The answers in my book gives 10 \frac{2}{3}

So if it is right then can someone enlighten me to what I have done wrong please.

Cheers

The Bob (2004 ©)

 

[Pseudo Statistic]

Hello all, again.

Simple question.

Compute \int_{9}^{4} (x - \frac{1}{\sqrt{x}}) dx

I did this:

\int_{9}^{4} (x - \frac{1}{x^{0.5}}) dx = \int_{9}^{4} (x - x^{-0.5}) dx

\left[ \frac{1}{2} x^2 - 2 x^{0.5} \right]_{9}^{4} = \left[ \frac{x^2}{2} - 2 \sqrt{x} \right]_{9}^{4}

(\frac{81}{2} - 6) - (\frac{16}{2} - 4) = \frac{81 - 12}{2} - \frac{16 - 8}{2} = \frac{69}{2} - \frac{8}{2} = \frac{61}{2} = 30.5

2 months of calculus and and still I am not sure of what I have done wrong. The answers in my book gives 10 \frac{2}{3}

So if it is right then can someone enlighten me to what I have done wrong please.

Cheers

The Bob (2004 ©)

Third step...
x - x^-0.5 gives 1/2x^2 - 2x^0.5... hmmm...
Your work looks right to me.. maybe a book error?
K.

 

[The Bob]

Third step...
x - x^-0.5 gives 1/2x^2 - 2x^0.5... hmmm...
Your work looks right to me.. maybe a book error?
K.

That is what I was hoping for. I know I make simply mistakes but now 4 people reckon it is right.

Cheers. Still though, can anyone else see anything wrong at all?

The Bob (2004 ©)

 

[dextercioby]

Nope,both your and the books answers are incorrect.My answer is -\frac{61}{2} and I'm sure of it,because even my old rusty Maple says it is so...

Daniel.

 

[Pseudo Statistic]

That's just weird.. even my calculator gives 30.5. :\
And yeah, I get the negative answer too... only mistake in his work I see now.

 

[The Bob]

Oh nuts. I see what I did wrong. I have written the original limtis of the question the wrong way around. My fault (obviously). I'm not used to the Tex of calculus problems.

It should have been \int_{4}^{9} (x - \frac{1}{\sqrt{x}}) dx to give \frac{61}{2}

Cheers. At least I can go to college on Wednesday and now I am right.

The Bob (2004 ©)

 

[dextercioby]

It seemed pretty weird to me too,with the larger value being the inferior limit,but,hey,anything is possible in mathemetics,right...?

Daiel.

 

[The Bob]

It seemed pretty weird to me too,with the larger value being the inferior limit,but,hey,anything is possible in mathemetics,right...?

Daiel.

Most things are possible but I am yet to see a larger, inferior limit too, Daiel.

The Bob (2004 ©)