JAMB 2002 · UME · Q48

Calculus (JAMB bridge)

If $\frac{dy}{dx} = 2x - 3$ and $y = 3$ when $x = 0$ , find $y$ in terms of $x$ .

Worked solution (try it first)

Integrate: $y = x^2 - 3x + c$ .
Put in $x = 0$ , $y = 3$ : $3 = 0 - 0 + c$ , so $c = 3$ .
So $y = x^2 - 3x + 3$ , option B.

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