a. We're looking for [tex]a[/tex] such that
[tex]\displaystyle\int_1^a\frac{\mathrm dx}{x^2}=\int_a^4\frac{\mathrm dx}{x^2}[/tex]
[tex]-\dfrac1x\bigg|_{x=1}^{x=a}=-\dfrac1x\bigg|_{x=a}^{x=4}[/tex]
[tex]1-\dfrac1a=\dfrac1a-\dfrac14[/tex]
[tex]\dfrac54=\dfrac2a\implies\boxed{a=\dfrac85}[/tex]
b. Integrating with respect to [tex]y[/tex] will make things easier.
[tex]y=\dfrac1{x^2}\implies x=\dfrac1{\sqrt y}[/tex]
(where we take the positive square root because we know [tex]x>0[/tex])
Now we want to find [tex]b[/tex] such that
[tex]\displaystyle\int_0^b\frac{\mathrm dy}{\sqrt y}=\int_b^1\frac{\mathrm dy}{\sqrt y}[/tex]
[tex]2\sqrt y\bigg|_{y=0}^{y=b}=2\sqrt y\bigg|_{y=b}^{b=1}[/tex]
[tex]2\sqrt b=2-2\sqrt b[/tex]
[tex]4\sqrt b=2\implies\boxed{b=\dfrac14}[/tex]
The area under a curve is function of the definite integral between the points under the curve.
(a) Find a
To do this, we make use of:
[tex]\mathbf{\int\limits^a_1 \frac{dx}{x^2} = \int\limits^4_a \frac{dx}{x^2}}[/tex]
Rewrite as:
[tex]\mathbf{\int\limits^a_1 x^{-2} \ dx = \int\limits^4_a x^{-2} \ dx }[/tex]
Integrate
[tex]\mathbf{ -x^{-1}|\limits^a_1 = -x^{-1}|\limits^4_a }[/tex]
Expand
[tex]\mathbf{ (-a^{-1}) -(-1^{-1}) = (-4^{-1}) -(-a^{-1})}[/tex]
[tex]\mathbf{ -\frac 1a +1 = -0.25 + \frac{1}{a} }[/tex]
Collect like terms
[tex]\mathbf{ \frac 1a + \frac{1}{a} =1 +0.25 }[/tex]
[tex]\mathbf{ \frac 1a + \frac{1}{a} =1.25 }[/tex]
Take LCM
[tex]\mathbf{ \frac{1+1}{a} =1.25 }[/tex]
[tex]\mathbf{ \frac{2}{a} =1.25 }[/tex]
Inverse both sides
[tex]\mathbf{ \frac{a}{2} =\frac{1}{1.25} }[/tex]
Multiply both sides by 2
[tex]\mathbf{ a =\frac{2}{1.25} }[/tex]
Simplify
[tex]\mathbf{ a =\frac{8}{5} }[/tex]
(b) Find b
In (a), we have:
[tex]\mathbf{y = \frac{1}{x^2}}[/tex]
Multiply both sides by x^2
[tex]\mathbf{x^2y = 1}[/tex]
Divide through by y
[tex]\mathbf{x^2 = \frac 1y}[/tex]
Take square roots
[tex]\mathbf{x = \frac {1}{\sqrt{y}}}[/tex]
So, the definite integral is:
[tex]\mathbf{\int\limits^b_0 \frac{dy}{\sqrt y} = \int\limits^1_b \frac{dy}{\sqrt y}}[/tex]
Rewrite as:
[tex]\mathbf{\int\limits^b_0 y^{-\frac 12}dy = \int\limits^1_b y^{-\frac 12}dy }[/tex]
Integrate
[tex]\mathbf{2y^{\frac 12}|\limits^b_0 = 2y^{\frac 12}|\limits^1_b}[/tex]
Expand
[tex]\mathbf{2(b^{\frac 12} - 0^{\frac 12}) = 2(1^{\frac 12} - b^{\frac 12})}[/tex]
[tex]\mathbf{2(b^{\frac 12} ) = 2(1 - b^{\frac 12})}[/tex]
Divide through by 2
[tex]\mathbf{b^{\frac 12} = 1 - b^{\frac 12}}[/tex]
Collect like terms
[tex]\mathbf{b^{\frac 12} + b^{\frac 12} = 1 }[/tex]
[tex]\mathbf{2b^{\frac 12} = 1 }[/tex]
Divide through by 2
[tex]\mathbf{b^{\frac 12} = \frac 12 }[/tex]
Square both sides
[tex]\mathbf{b = \frac 14 }[/tex]
Hence, the value of b is 1/4
Read more about areas under curves at:
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