Challenging +1.2 This is a structured multi-part question that guides students through standard trapezium rule application, generalisation to n strips, and connects to limits and integration. While part (e) requires recognising the derivative definition, the heavy scaffolding and routine techniques make it moderately above average difficulty but not exceptionally challenging for AEA level.
4.(a)Use the trapezium rule with 4 strips to find an approximate value for
$$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$
(b)Use the trapezium rule with \(n\) strips to write down an expression that would give an approximate value for
$$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$
(c)Hence show that
$$\int _ { 0 } ^ { 1 } 16 ^ { x } \mathrm {~d} x = \lim _ { n \rightarrow \infty } \left( \frac { 1 } { n } \left( 1 + 16 ^ { \frac { 1 } { n } } + \ldots + 16 ^ { \frac { n - 1 } { n } } \right) \right)$$
(d)Use integration to determine the exact value of
$$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$
Given that the limit exists,
(e)use part(c)and the answer to part(d)to determine the exact value of
$$\lim _ { x \rightarrow 0 } \frac { 16 ^ { x } - 1 } { x }$$
4.(a)Use the trapezium rule with 4 strips to find an approximate value for
$$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$
(b)Use the trapezium rule with $n$ strips to write down an expression that would give an approximate value for
$$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$
(c)Hence show that
$$\int _ { 0 } ^ { 1 } 16 ^ { x } \mathrm {~d} x = \lim _ { n \rightarrow \infty } \left( \frac { 1 } { n } \left( 1 + 16 ^ { \frac { 1 } { n } } + \ldots + 16 ^ { \frac { n - 1 } { n } } \right) \right)$$
(d)Use integration to determine the exact value of
$$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$
Given that the limit exists,\\
(e)use part(c)and the answer to part(d)to determine the exact value of
$$\lim _ { x \rightarrow 0 } \frac { 16 ^ { x } - 1 } { x }$$
\hfill \mbox{\textit{Edexcel AEA 2023 Q4 [16]}}