WAEC - Further Mathematics (2013 - No. 14)
Given that \(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}\), find y.
\(2x^{\frac{3}{2}} + c\)
\(\frac{2}{3}x^{\frac{3}{2}} + c\)
\(\frac{3}{2}x^{\frac{3}{2}} + c\)
\(\frac{2}{3}x^{2} + c\)
Explicação
\(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x} = x^{\frac{1}{2}}\)
\(y = \int x^{\frac{1}{2}} \mathrm {d} x\)
= \(\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + c = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c \)
= \(\frac{2}{3}x^{\frac{3}{2}} + c\)