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IgorC [24]
3 years ago
11

Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Mathematics
1 answer:
Arlecino [84]3 years ago
6 0
y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt

By the fundamental theorem of calculus,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}

Now the arc length over an arbitrary interval (a,b) is

\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx

But before we compute the integral, first we need to make sure the integrand exists over it. x^{3/2} is undefined if x, so we assume a\ge0 and for convenience that a. Then

\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)
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