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S is equal. . ut 1/2 at2 with calculus method...........no need silly answer​

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We have to DERIVE s = UT + ½at² by calculus method

CONSIDER the VELOCITY - Displacement Relation :

\sf \: v = \dfrac{dx}{dt}

From the First Kinematic Equation,

\sf \: v = u + at

Thus,

\sf \: u + at = \dfrac{dx}{dt} \\ \\ \longrightarrow \: \sf \: (u + at)dt = dx

Integrating on both sides,

\longrightarrow \displaystyle \: \sf \int _0^s{x}^{0} .ds \: = \int_0^tu {t}^{0}.dt + \int_0^tat.dt \\ \\ \longrightarrow \: \displaystyle \: \sf \: \int _0^s{x}^{0} .ds \: =u \int_0^t {t}^{0}.dt + a\int_0^tt.dt \\ \\ \longrightarrow \: \sf \: \bigg[x \bigg]_0^s = u\bigg[t \bigg]_0^t + a\bigg[ \dfrac{ {t}^{2} }{2} \bigg]_0^t \\ \\ \longrightarrow \: \sf \: (s - 0) = u(t - 0) + \dfrac{a}{2} ( {t}^{2} - {0}^{2} ) \\ \\ \large{ \longrightarrow \: \boxed{ \boxed{ \sf \: s = ut + \dfrac{1}{2} a {t}^{2} }}}


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