Finding the area under a curve.

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joshmeyer

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I'm given the equations
[math]-7x^2[/math]and
[math]-√x[/math]I know that the intersection points are where the 2 equations equal each other, but for the life of me I can't recall how to solve
[math]-7x^2=-√x[/math]Once I figure that out I simply do this math, right?
[math]-\frac{7}{3}x^3-(-\frac{2}{3}x^\frac{3}{2})[/math]Subtituting the values of intersection for x, and subtracting the smaller value from the larger value.
 
joshmeyer said:
I'm given the equations
[math]-7x^2[/math]and
[math]-√x[/math]I know that the intersection points are where the 2 equations equal each other, but for the life of me I can't recall how to solve
[math]-7x^2=-√x[/math]Once I figure that out I simply do this math, right?
[math]-\frac{7}{3}x^3-(-\frac{2}{3}x^\frac{3}{2})[/math]Subtituting the values of intersection for x, and subtracting the smaller value from the larger value.
Click to expand...
Hi Josh,
I would square both sides as follows:
[math](-7x^2)^2=(-\sqrt{x})^2\\ 49x^4=x\\ 49x^4-x=0\\[/math]Hopes this help
:)
 
You say that you have two equations, what are they. Equations have equal signs.
 
Ok I'm still very stumped.
1650993947324.png
I don't even have the faintest idea of how to simplify the left side of that mess.
 
joshmeyer said:
Ok I'm still very stumped.
View attachment 32400
I don't even have the faintest idea of how to simplify the left side of that mess.
Click to expand...
Algebraic mistake when finding the intercepts. You have a plus when it should be a minus.
[math]49x^4=x\\ 49x^4\red{-}x=0\\ x(49x^3\red{-}1)=0\\ x=0 ,\, x=\frac{1}{\sqrt[3]{49}}=\frac{1}{7^{2/3}}[/math]
 
BigBeachBanana said:
Algebraic mistake when finding the intercepts. You have a plus when it should be a minus.
[math]49x^4=x\\ 49x^4\red{-}x=0\\ x(49x^3\red{-}1)=0\\ x=0 ,\, x=\frac{1}{\sqrt[3]{49}}=\frac{1}{7^{2/3}}[/math]
Click to expand...
Okay, yes, but I still have no idea how you went from
1650994985238.png
to
1650995005150.png
Edit: I feel, so, so so, so, so dumb. I see it now
 
Last edited:
joshmeyer said:
Okay, yes, but I still have no idea how you went from
View attachment 32401
to
View attachment 32402
Click to expand...
I'm not doing anything to the integration yet, only the intercept.
[math]\frac{1}{\sqrt[3]{49}}=\frac{1}{\sqrt[3]{7{^2}}}=\frac{1}{7^{2/3}}=7^{-2/3}[/math]
For the integral, now you have:

[math]\int_{0}^{7^{-2/3}}-7x^2+\sqrt{x}\,dx= \frac{-7x^3}{3}+\frac{2x^{3/2}}{3}\biggr\rvert_{0}^{7^{-2/3}}= \frac{-7{(7^{-2/3}})^3}{3}+\frac{2({7^{-2/3}})^{3/2}}{3}-0[/math]Can you simplify?
 
Last edited:
To rationalize \(\displaystyle \dfrac{1}{\sqrt[3]7}\) you simple do the following.

\(\displaystyle \dfrac{1}{\sqrt[3]7} = \dfrac{1}{\sqrt[3]7}*\dfrac{\sqrt[3]7\sqrt[3]7}{\sqrt[3]7\sqrt[3]7}=\dfrac{\sqrt[3]49}{7}\)
 
Steven G said:
To rationalize \(\displaystyle \dfrac{1}{\sqrt[3]7}\) you simple do the following.

\(\displaystyle \dfrac{1}{\sqrt[3]7} = \dfrac{1}{\sqrt[3]7}*\red{\dfrac{\sqrt[3]7\sqrt[3]7}{\sqrt[3]7\sqrt[3]7}}=\dfrac{\sqrt[3]49}{7}\)
Click to expand...
Hi Steven, I think you might've made an algebraic mistake.
\(\displaystyle \dfrac{1}{\sqrt[3]7}\neq \dfrac{\sqrt[3]49}{7}\)

I think you meant:
\(\displaystyle \dfrac{1}{\sqrt[3]7} = \dfrac{1}{\sqrt[3]7}*\dfrac{\sqrt[3]7}{\sqrt[3]7}=\dfrac{\sqrt[3]{7}}{7}\)
 
BigBeachBanana said:
Hi Steven, I think you might've made an algebraic mistake.
\(\displaystyle \dfrac{1}{\sqrt[3]7}\neq \dfrac{\sqrt[3]49}{7}\)

I think you meant:
\(\displaystyle \dfrac{1}{\sqrt[3]7} = \dfrac{1}{\sqrt[3]7}*\dfrac{\sqrt[3]7}{\sqrt[3]7}=\dfrac{\sqrt[3]{7}}{7}\)
Click to expand...
@Steven G
Ignore post#11. Realized you're rationalizing a different quantity than the OP.
 
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