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STEP I 1992 question 8 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1992 question 8 solution


[Unparseable or potentially dangerous latex formula. Error 6 ]

= \frac{\text{d}}{\text{d} x} [\int f(x) - \int f(a) ]

= \frac{\text{d}}{\text{d} x}(\int f(x)) = f(x)

Consider

[Unparseable or potentially dangerous latex formula. Error 6 ].

You can just differentiate both sides from here and get

f'(x) = f(x) \Rightarrow f(x) = Ae^x

and plug this into the integrals in the second and third part:

[Unparseable or potentially dangerous latex formula. Error 6 ]

[Unparseable or potentially dangerous latex formula. Error 6 ]

Ae^x = Ae^x - A + 1 \Rightarrow A = 1


[Unparseable or potentially dangerous latex formula. Error 6 ]

[Unparseable or potentially dangerous latex formula. Error 6 ]

Ae^x = Ae^x - A \Rightarrow A=0


We wish to solve \displaystyle \int_0^x f(t)\,\text{d} t = \int_x^1 t^2f(t)\text{d} t +x - \frac{x^5}{5} + C. As before, differentiate w.r.t. x:

f(x) = -x^2f(x) + (1-x^4) \implies (1+x^2)f(x) = (1-x^4) = (1-x^2)(1+x^2) and so f(x) = 1-x^2.

Substitute back in: \displaystyle \int_0^x (1-t^2)\,dt = \int_x^1 t^2-t^4dt +x - \frac{x^5}{5} + C

\displaystyle x-\frac{x^3}{3} = \left[\frac{t^3}{3} - \frac{t^5}{5}\right]_x^1 +x - \frac{x^5}{5} + C

\displaystyle x-\frac{x^3}{3} = \frac{1}{3}-\frac{1}{5} - \frac{x^3}{3} + \frac{x^5}{5} +x - \frac{x^5}{5} + C

Deduce C = \frac{1}{5}-\frac{1}{3} = -\frac{2}{15}.

Solution by insparato and DFranklin.

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