Additional Mathematics 0606 / Calculus — Integration / 1.611.61: Definite integral leading to an exponential equation0606/13/M/J/17 — Question 9 · 8 marksMark as done · Save for later · Show all solutionsIt is given that ∫−kk(15e5x−5e−5x) dx=6\displaystyle\int_{-k}^{k} (15\mathrm{e}^{5x} - 5\mathrm{e}^{-5x})\,\mathrm{d}x = 6∫−kk(15e5x−5e−5x)dx=6.(i)Show that e5k−e−5k=3\mathrm{e}^{5k} - \mathrm{e}^{-5k} = 3e5k−e−5k=3.[5]▸ Answer(ii)Hence, using the substitution y=e5ky = \mathrm{e}^{5k}y=e5k, or otherwise, find the value of kkk.[3]▸ Answer▸ Official mark scheme← 1.33: Integrating an exponential and finding a second derivative1.62: Product rule and integration of a logarithm →
