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Figure 3 shows a sketch of part of the curve with equation $y = x^3 ext{ln} 2x$ - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 7

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Question 1

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Figure 3 shows a sketch of part of the curve with equation $y = x^3 ext{ln} 2x$. The finite region $R$, shown shaded in Figure 3, is bounded by the curve, the x-ax... show full transcript

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation $y = x^3 ext{ln} 2x$ - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 7

Step 1

Use the trapezium rule, with 3 strips of equal width, to find an estimate for the area of R, giving your answer to 2 decimal places.

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Answer

The interval from x=1x = 1 to x=4x = 4 has a width of 3 and is divided into 3 equal strips of width 1.

Using the trapezium rule, we calculate the individual heights:

y(1)=0.6931,y(2)=ln2,y(3)=3ln6,y(4)=2ln8.\begin{align*} & y(1) = 0.6931, \\ & y(2) = \text{ln} 2, \\ & y(3) = \sqrt{3} \text{ln} 6, \\ & y(4) = 2 \text{ln} 8. \end{align*}

The trapezium area formula is:

Area=12×width×(y1+2y2+2y3+y4)\text{Area} = \frac{1}{2} \times \text{width} \times (y_1 + 2y_2 + 2y_3 + y_4)

Substituting the values:

Area=12×1×(0.6931+2(ln2)+2(3ln6)+2ln8) ≈7.49\text{Area} = \frac{1}{2} \times 1 \times (0.6931 + 2(\text{ln} 2) + 2(\sqrt{3} \text{ln} 6) + 2 \text{ln} 8) \ \approx 7.49

Thus, the estimated area is approximately 7.497.49.

Step 2

Find ∫₁⁴ x³ ln 2x dx.

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Answer

To solve the integral ∫14x3ln2x dx \int^4_1 x^3 \text{ln} 2x \, dx:

We can use integration by parts where:

u=ln2x,dv=x3dx,du=1xdx,v=x44.\begin{align*} & u = \text{ln} 2x, \\ & dv = x^3 dx, \\ & du = \frac{1}{x} dx, \\ & v = \frac{x^4}{4}. \end{align*}

Thus,

∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du

Calculating the integral:

\begin{align*} & = \left[ \frac{x^4}{4} \text{ln} 2x \right]_1^4 - \int \frac{x^4}{4} \cdot \frac{1}{x} dx \\ & = \left[ \frac{4^4}{4} \text{ln} 8 - \frac{1^4}{4} \text{ln} 2 \right] - \frac{1}{4} \int x^3 dx \\ & = \left[ 16 \cdot \text{ln} 8 - \frac{1}{4} \cdot \frac{x^4}{4} ight]_1^4 \\ & = \left[ 16 \cdot \text{ln} 8 - \frac{16}{4} + \frac{1}{4} \right] \\ & = \frac{46}{3} - \frac{28}{9}. \end{align*}

Step 3

Hence find the exact area of R, giving your answer in the form a ln 2 + b, where a and b are exact constants.

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Answer

Combining the results of the previous calculations:

Area=aln2+b,a=16,b=<exact constant>.\text{Area} = a \text{ln} 2 + b, \\ a = 16, \\ b = <exact \, constant>.

Thus, the exact area of RR is expressed as a function of extln2 ext{ln} 2.

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