C2 2017 3 Edexcel

(a)

When x = 1 we have:

y = 5^1 + \log_2 (1 + 1) = 5 + \log_2 2 = 5 + 1 = 6

(b)

The step length is 0.5, so by the trapezium rule we have:

\begin{align*}\int_0^2 (5^x + \log_2 (x + 1)) \mathrm dx & \approx \frac {0.5} 2 (1 + 26.585 + 2(2.821 + 6 + 12.502)) \\ & = 17.55775 \\ & = 17.56 \text { (2dp)}\end{align*}

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We have:

\displaystyle \int_0^2 (5 + 5^x + \log_2 (x + 1)) \mathrm dx = \int_0^2 5 \mathrm dx + \int_0^2 (5^x + \log_2 (x + 1)) \mathrm dx

We’ve already found that:

\displaystyle \int_0^2 (5^x + \log_2 (x + 1)) \mathrm dx \approx 17.56

and we have:

\displaystyle \int_0^2 5 \mathrm dx = 5(2 - 0) = 10

So:

\displaystyle \int_0^2 (5 + 5^x + \log_2 (x + 1)) \mathrm dx \approx 27.56

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