5x^3+40= the integral of f(t) dt from a to x solve for a

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15-02-2017
Mathematics

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5x^3+40= the integral of f(t) dt from a to x solve for a

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LammettHash LammettHash · Answer 1 · 2017-02-19T23:09:13+01:00

[tex]5x^3+40=\displaystyle\int_a^xf(t)\,\mathrm dt[/tex]

Differentiate both sides with respect to [tex]x[/tex]:

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[5x^3+40\right]=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_a^xf(t)\,\mathrm dt[/tex]
[tex]15x^2=f(x)[/tex]

This means

[tex]\displaystyle\int_a^x15t^2\,\mathrm dt=5x^3+40[/tex]
[tex]5t^3\bigg|_{t=a}^{t=x}=5x^3+40[/tex]
[tex]5(x^3-a^3)=5x^3+40[/tex]
[tex]-5a^3=40[/tex]
[tex]a^3=-8[/tex]
[tex]a=-2[/tex]