Solution — 2016 Question 3 Graphs| Differentiation & Applications| Integration & Applications

$$\begin{array}{cc}& \text{Gradient}\\ & =\frac{\mathrm{d}y}{\mathrm{d}x}\\ & =-1.6065\\ & =-1.61∎\text{ (to 3 s.f.)}\end{array}$$

$$\begin{array}{cc}y& ={\mathrm{e}}^{-x}-x^2\\ & ={\mathrm{e}}^{-\left(0.5\right)}-{0.5}^2\\ & =0.35653\end{array}$$

$$\begin{array}{cc}y-y_1& =m(x-x_1)\\ y-0.35653& =-1.6065(x-0.5)\\ & =-1.6065x+0.80327\\ y& =-1.6065x+1.1598∎\end{array}$$

$$\begin{array}{cc}& {\displaystyle {\int }_0^k({\mathrm{e}}^{-x}-x^2)\mathrm{d}x}\\ & ={[-{\mathrm{e}}^{-x}-\frac{x^3}{3}]}_0^k\\ & =-{\mathrm{e}}^{-k}-\frac{k^3}{3}-(-{\mathrm{e}}^{-\left(0\right)}-\frac{0^3}{3})\\ & =-{\mathrm{e}}^{-k}-\frac{k^3}{3}+1∎\end{array}$$