Mixture of vectors and calculus

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The range of the projectile is given by $R= \dfrac{u^2 \sin 2\theta }{g}$

Which is , in this case, $R = \dfrac{5 \sin 90^\circ}{10} = \dfrac{1}{2}$

Thus the equation of the trajectory, which is $y=x\tan\theta \biggl( 1 - \dfrac{x}{R}\biggr)$ , is in this case

$y = x \tan 45^\circ \biggl( 1- \dfrac{x}{\frac{1}{2}} \biggr) \\ \quad= x \times 1 \times (1-2x) \\ \quad = x-2x^2$

Area under this curve is given by

$\displaystyle \int_0 ^\frac{1}{2} y dx = \int_0 ^\frac{1}{2} (x -2x^2) dx \\ \quad\quad= \dfrac{x^2}{2} - \dfrac{2x^3}{3} \Bigg| _0 ^2 \\ \quad\quad= \dfrac{1}{8} - \dfrac{2}{3}\times \dfrac{1}{8}\\ \quad\quad = \dfrac{1}{3\times 8 }\\ \quad\quad = \dfrac{1}{24}$

Thus answer is $\boxed{24}$