Need help with two questions (Integration) (1 Viewer)

BlueGas · Feb 9, 2015

I need help with questions 4 and 6 from the picture below, thanks.

turnerloos · Feb 9, 2015

i broke my neck reading it.

BlueGas · Feb 9, 2015

turnerloos said:
i broke my neck reading it.

It's not like as if you're reading an essay, it's only two questions.

Silly Sausage · Feb 9, 2015

Q4.
$y=\frac{1}{x^2}$
becomes
$\frac{1}{\sqrt{y}}$
Thus,
$\int^{2}_{1}\frac{1}{\sqrt{y}} \, dy$
self explanatory from there on;

Q6.
A1,
$\int^{2}_{1}x^2 \, dx$
=
$[\frac{x^3}{3}]^{2}_{1}=\frac{7}{3}$
A2,
$x^2=y$
=
$x=\sqrt{y}$
=
$\int^{4}_{1}{\sqrt{y} }\, dy$
=
$[\frac{2(y^{\frac{5}{3}})}{3}]^4_1$
= $\frac{14}{3}$
$\frac{14}{3}+\frac{7}{3}=7$

BlueGas · Feb 9, 2015

Silly Sausage said:
Q4.
$y=\frac{1}{x^2}$
becomes
$\frac{1}{\sqrt{y}}$
Thus,
$\int^{2}_{1}\frac{1}{\sqrt{y}} \, dy$
self explanatory from there on;

Q6.
A1,
$\int^{2}_{1}x^2 \, dx$
=
$[\frac{x^3}{3}]^{2}_{1}=\frac{7}{3}$
A2,
$x^2=y$
=
$x=\sqrt{y}$
=
$\int^{4}_{1}{\sqrt{y} }\, dy$
=
$[\frac{2(y^{\frac{5}{3}})}{3}]^4_1$
= $\frac{14}{3}$
$\frac{14}{3}+\frac{7}{3}=7$

Thanks man, one question though, how would you integrate in Q4 after you make x the subject? Also in question 6, I thought A1 area was only the square A1 was in, not the whole shaded area, thanks for clearing it up.

BlueGas · Feb 9, 2015

Also why did you do dx in A1 and dy in A2? Why didn't you just make x the subject in both areas?

Silly Sausage · Feb 9, 2015

I did dy because we are integrating with respect to y.
You need to do this in order to find the area along the y-axis
$\int^{2}_{1}y^\frac{-1}{2}\, dy$
=
$\frac{y^\frac{-1}{2}}{\frac{1}{2}}$
=
$[2\sqrt{y} ]^2_1$

LostInHSC · Feb 9, 2015

BlueGas said:
Thanks man, one question though, how would you integrate in Q4 after you make x the subject? Also in question 6, I thought A1 area was only the square A1 was in, not the whole shaded area, thanks for clearing it up.

$\int \frac{1}{\sqrt{y}} = \int (y)^{-\frac{1}{2}} = \frac{y^{\frac{1}{2}}}{\frac{1}{2}\times1} + C = 2\sqrt{y} + C$

BlueGas · Feb 9, 2015

LostInHSC said:
$\int \frac{1}{\sqrt{y}} = \int (y)^{-\frac{1}{2}} = \frac{y^{\frac{1}{2}}}{\frac{1}{2}\times1} + C = 2\sqrt{y} + C$

Silly Sausage said:
I did dy because we are integrating with respect to y.
You need to do this in order to find the area along the y-axis
$\int^{2}_{1}y^\frac{-1}{2}\, dy$
=
$\frac{y^\frac{-1}{2}}{\frac{1}{2}}$
=
$[2\sqrt{y} ]^2_1$

Okay thanks guys.

LostInHSC · Feb 10, 2015

Silly Sausage said:
Q4.
$y=\frac{1}{x^2}$
becomes
$\frac{1}{\sqrt{y}}$
Thus,
$\int^{2}_{1}\frac{1}{\sqrt{y}} \, dy$
self explanatory from there on;

Q6.
A1,
$\int^{2}_{1}x^2 \, dx$
=
$[\frac{x^3}{3}]^{2}_{1}=\frac{7}{3}$
A2,
$x^2=y$
=
$x=\sqrt{y}$
=
$\int^{4}_{1}{\sqrt{y} }\, dy$
=
$[\frac{2(y^{\frac{5}{3}})}{3}]^4_1$
= $\frac{14}{3}$
$\frac{14}{3}+\frac{7}{3}=7$

Sorry just confused: Shouldn't we multiply your answer for question 4 by 2 (taking into account area bounded by the y-axis in quadrant 1 and quadrant 2)?

InteGrand · Feb 10, 2015

LostInHSC said:
Sorry just confused: Shouldn't we multiply your answer for question 4 by 2 (taking into account area bounded by the y-axis in quadrant 1 and quadrant 2)?

Yes, the area is double that integral.

Need help with two questions (Integration) (1 Viewer)

BlueGas

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