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Daniel [21]
3 years ago
13

Find the number a such that the line x = a bisects the area under the curve y = 1/x2 for 1 ≤ x ≤ 4. 8 5​ (b) find the number b s

uch that the line y = b bisects the area in part (a).
Mathematics
1 answer:
IceJOKER [234]3 years ago
3 0

a. We're looking for a such that

\displaystyle\int_1^a\frac{\mathrm dx}{x^2}=\int_a^4\frac{\mathrm dx}{x^2}

-\dfrac1x\bigg|_{x=1}^{x=a}=-\dfrac1x\bigg|_{x=a}^{x=4}

1-\dfrac1a=\dfrac1a-\dfrac14

\dfrac54=\dfrac2a\implies\boxed{a=\dfrac85}

b. Integrating with respect to y will make things easier.

y=\dfrac1{x^2}\implies x=\dfrac1{\sqrt y}

(where we take the positive square root because we know x>0)

Now we want to find b such that

\displaystyle\int_0^b\frac{\mathrm dy}{\sqrt y}=\int_b^1\frac{\mathrm dy}{\sqrt y}

2\sqrt y\bigg|_{y=0}^{y=b}=2\sqrt y\bigg|_{y=b}^{b=1}

2\sqrt b=2-2\sqrt b

4\sqrt b=2\implies\boxed{b=\dfrac14}

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