C4 2017 3 Edexcel

(a)

y = \dfrac 6 {e^{0.2} + 2} = 1.862542\ldots = 1.86254 \text { (5dp)}

(b)

The step length here is 0.2, so we have:

\begin{align*} \mathrm{area}(R) &\approx \frac 1 2 \times 0.2 \left(2 + 1.27165 + 2(1.86254 + 1.71830 + 1.56981 + 1.41994)\right) \\ & = 1.641283 \\ & = 1.6413 \text { (4dp)}\end{align*}

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The area of R is equal to the integral:

\displaystyle \int_0^1 \frac 6 {e^x + 2} \mathrm dx

If u = e^x then \displaystyle \frac {\mathrm dx} {\mathrm du} = e^{-x} = \frac 1 u. When x = 0, u = a = e^0 = 1 and when x = 1, u = b = e^1 = e. So:

\begin{align*}\int_0^1 \frac 6 {e^x + 2} \mathrm dx & = \int_1^e \frac 6 {u + 2} \times \frac 1 u \mathrm du \\ & = \int_1^e \frac 6 {u(u + 2)} \mathrm du\end{align*}

(d)

We want to use partial fractions. If:

\displaystyle \frac 6 {u(u + 2)} = \frac A u + \frac B {u + 2}

Then for u \ne 0, -2 we have:

6 = A(u + 2) + Bu

So 2A = 6, giving A = 3. And A + B = 0, so B = -3, giving:

\displaystyle \frac 6 {u(u + 2)} = \frac 3 u - \frac 3 {u + 2}

We therefore have:

\begin{align*}\int_1^e \frac 6 {u(u+2)} \mathrm du & = 3 \int_1^e \left(\frac 1 u - \frac 1 {u + 2}\right) \mathrm du \\ & = 3 \left[\ln u - \ln(u + 2)\right]_1^e \\ & = 3 \left(\ln e - \ln(e + 2) - \ln 1 + \ln 3\right) \\ & = 3 \left(1 + \ln \left(\frac 3 {e + 2}\right)\right) \\ & = 3 + 3 \ln \left(\frac 3 {e + 2}\right)\end{align*}

Other forms are possible.

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