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1.32 Integral associated with the parabola

(i) For the parabola, we substitute $u=ax+b$ in

\int_{0}^{t}(ax+b)^{1/2}dx.

Then $du=a\,dx$ , so $dx=\frac{1}{a}du$ . Hence we obtain

\int_{0}^{t}(ax+b)^{\frac{1}{2}}\,dx=\,{\int_{b}^{at+b}u^{\frac{1}{2}}\cdot% \frac{1}{a}\,du}\,{=\left[\frac{2}{3a}u^{\frac{3}{2}}\right]^{at+b}_{b}}