Basic Calculus II Integral Questions - Riemann Sums, Absolute Integrals, etc.

[ardentmed]

Hey guys, I'd appreciate some help for this problem set I'm working on currently

The u-substitution for the first one is somewhat tricky. I ended up getting 1/40(u)^5/2 - 2 (u) ^3/2 +C, which I'm not too sure about. I took u from radical 3+2x^4.

For the second question, I split the integral into two parts, one from 3 to 1, and the other from 1 to -2, and used a negative value for the latter, since x < 1 is negative. This gave me 20/3 as a final answer.

Moreover, for the third question, I used a trigonometric identity u substitution and got \pi^2 / 72.

For the last question (the word problem), I got -3.33m for displacement and took the absolute for velocity to get speed, which gave me 218/3 as the final answer.

Also, for these questions:

I'm confused as to how these work. For instance, I'm assuming that the FTC must be used, where the formula used would be f(b)-f(a), correct?
If so, I got 4x^3 radical(2-x^4) -2 for the first one. The second one is 3x^2 (sin(x^3)) + sincosx * cosx.

Also for the last two, I got 3/5x^5/4 -2x + 3x^1/3 +c from multiplying the values in the bracket by themselves (expansion). Also, I got 2 for the final one, since sinx > 0 is positive, and sinx < 0 is negative, although I'm not too sure about that. Thanks in advance, guys. I really appreciate it.

 

[ineedhelpnow]

for the first one you can do:

$u=x^4$
$du=4x^3dx$
$dx=\frac{1}{4x^3}du$

$\int \ x^7 \sqrt{3+2u}*\frac{1}{4x^3}du$
$\int \ \frac{x^4 \sqrt{3+2u}}{4}du$
$\int \ \frac{u \sqrt{3+2u}}{4}du$
$\frac{1}{4} \int \ u \sqrt{3+2u}du$

$v=3+2u$
$dv=2du$
$du=\frac{1}{2}dv$

set the whole thing in terms of v and then plug 3+2u in for v and u=x^4. i suppose there might be another way to do it but that's the only way i can think of.

i got the same answer for the third one (b on the first attachment) using sin for the substitution.

 

[ineedhelpnow]

your part a for the word problem is correct but for part b:

$v(t)=(t-4)(t+2)$
$\int_{1}^{4} \ 8+2t-t^2 dt + \int_{4}^{6} \ t^2-2t-8 dt$
and that is equal to 98/3

 

[ineedhelpnow]

the last two are correct as well, only i think you meant $\frac{3x^{5/3}}{5}-2x+3^{1/3}$

 

[ardentmed]

the last two are correct as well, only i think you meant $\frac{3x^{5/3}}{5}-2x+3^{1/3}$

Thanks a ton, I really appreciate the help and prompt reply. I'll go through the last one again just to make sure.

 

[ineedhelpnow]

no problem. i would try to work out the other ones also but i don't have time. i have to get ready for my own calc 2 test which is like in an hour. hopefully answering your question can work as practice for me :D

 

[Prove It]

Hey guys, I'd appreciate some help for this problem set I'm working on currently

The u-substitution for the first one is somewhat tricky. I ended up getting 1/40(u)^5/2 - 2 (u) ^3/2 +C, which I'm not too sure about. I took u from radical 3+2x^4.

For the second question, I split the integral into two parts, one from 3 to 1, and the other from 1 to -2, and used a negative value for the latter, since x < 1 is negative. This gave me 20/3 as a final answer.

Moreover, for the third question, I used a trigonometric identity u substitution and got \pi^2 / 72.

For the last question (the word problem), I got -3.33m for displacement and took the absolute for velocity to get speed, which gave me 218/3 as the final answer.

Also, for these questions:

I'm confused as to how these work. For instance, I'm assuming that the FTC must be used, where the formula used would be f(b)-f(a), correct?
If so, I got 4x^3 radical(2-x^4) -2 for the first one. The second one is 3x^2 (sin(x^3)) + sincosx * cosx.

Also for the last two, I got 3/5x^5/4 -2x + 3x^1/3 +c from multiplying the values in the bracket by themselves (expansion). Also, I got 2 for the final one, since sinx > 0 is positive, and sinx < 0 is negative, although I'm not too sure about that. Thanks in advance, guys. I really appreciate it.

In the second question, splitting up your integral is the right approach. But what are the roots of $\displaystyle \begin{align*} x^2 - 1 \end{align*}$?

Your answer to the third question is correct.

In the fourth problem, are you given the position of the particle at any point in time?

 

[Prove It]

Hey guys, I'd appreciate some help for this problem set I'm working on currently

The u-substitution for the first one is somewhat tricky. I ended up getting 1/40(u)^5/2 - 2 (u) ^3/2 +C, which I'm not too sure about. I took u from radical 3+2x^4.

For the second question, I split the integral into two parts, one from 3 to 1, and the other from 1 to -2, and used a negative value for the latter, since x < 1 is negative. This gave me 20/3 as a final answer.

Moreover, for the third question, I used a trigonometric identity u substitution and got \pi^2 / 72.

For the last question (the word problem), I got -3.33m for displacement and took the absolute for velocity to get speed, which gave me 218/3 as the final answer.

Also, for these questions:

I'm confused as to how these work. For instance, I'm assuming that the FTC must be used, where the formula used would be f(b)-f(a), correct?
If so, I got 4x^3 radical(2-x^4) -2 for the first one. The second one is 3x^2 (sin(x^3)) + sincosx * cosx.

Also for the last two, I got 3/5x^5/4 -2x + 3x^1/3 +c from multiplying the values in the bracket by themselves (expansion). Also, I got 2 for the final one, since sinx > 0 is positive, and sinx < 0 is negative, although I'm not too sure about that. Thanks in advance, guys. I really appreciate it.

$\displaystyle \begin{align*} \int{ \left( \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \right) ^2 \, \mathrm{d}x } &= \int{ \left[ \frac{ \left( \sqrt[3]{x} \right) ^2 - 1}{\sqrt[3]{x}} \right] ^2 \, \mathrm{d}x } \\ &= \int{ \frac{ \left[ \left( \sqrt[3]{x} \right) ^2 - 1 \right] ^2 }{ \left( \sqrt[3]{x} \right) ^2 } \,\mathrm{d}x } \\ &= 3 \int{ \left[ \left( \sqrt[3]{x} \right) ^2 - 1 \right] ^2 \, \frac{1}{3\left( \sqrt[3]{x} \right) ^2 } \,\mathrm{d}x } \end{align*}$

Now notice that $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt[3]{x} \right) = \frac{\mathrm{d}}{\mathrm{d}x} \left( x^{\frac{1}{3}} \right) = \frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3 \left( \sqrt[3]{x} \right) ^2 } \end{align*}$, so the substitution $\displaystyle \begin{align*} u = \sqrt[3]{x} \implies \mathrm{d}u = \frac{1}{3\left( \sqrt[3]{x} \right) ^2 } \,\mathrm{d}x \end{align*}$ is appropriate...

 

[ardentmed]

In the second question, splitting up your integral is the right approach. But what are the roots of x2−1?

Your answer to the third question is correct.

In the fourth problem, are you given the position of the particle at any point in time?

Oh, I see. So because the roots are both +1 and -1, the integral, in actuality, is split up into four parts since both roots are part of the domain. Is that correct? So would it be positive from 3 to 1 and negative from 1 to -1, and then positive again from -1 to -2? That's what's confusing me at the moment.

Thanks again.

 

[Prove It]

Oh, I see. So because the roots are both +1 and -1, the integral, in actuality, is split up into four parts since both roots are part of the domain. Is that correct? So would it be positive from 3 to 1 and negative from 1 to -1, and then positive again from -1 to -2? That's what's confusing me at the moment.

Thanks again.

Yes that is correct (so really it's split into three parts, not four...)