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Figure 2 shows a sketch of the curve with equation $y = x^3 ext{ ln}(x^2 + 2)$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 5
Question 6
Figure 2 shows a sketch of the curve with equation $y = x^3 ext{ ln}(x^2 + 2)$, $x > 0$. The finite region $R$, shown shaded in Figure 2, is bounded by the curve, ... show full transcript
Worked Solution & Example Answer:Figure 2 shows a sketch of the curve with equation $y = x^3 ext{ ln}(x^2 + 2)$, $x > 0$ - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 5
Step 1
Complete the table above giving the missing values of $y$ to 4 decimal places.
Answer
To find the missing values of , we compute:
-
For x = rac{ oot{2}}{4}:
oot{2}}{4} igg)^3 ext{ ln} igg( igg( rac{ oot{2}}{4} igg)^2 + 2 igg)$$
After evaluating, we obtain:
ightarrow 0.3240$$ 2. For $x = rac{3 oot{2}}{4}$: $$y = igg( rac{3 oot{2}}{4} igg)^3 ext{ ln} igg( igg( rac{3 oot{2}}{4} igg)^2 + 2 igg)$$ This evaluates to: $$y ightarrow 1.3596$$ 3. For $x = rac{ oot{2}}{2}$: $$y = igg( rac{ oot{2}}{2} igg)^3 ext{ ln} igg( igg( rac{ oot{2}}{2} igg)^2 + 2 igg)$$ This results in: $$y ightarrow 3.9210$$Step 2
Use the trapezium rule, with all the values of $y$ in the completed table, to obtain an estimate for the area of $R$, giving your answer to 2 decimal places.
Answer
Applying the trapezium rule, we calculate:
oot{2}}{4} igg [ y_0 + 2y_1 + 2y_2 + y_3 igg ]$$ Substituting the values from the table: - $y_0 = 0$ - $y_1 = 0.3240$ - $y_2 = 1.3596$ - $y_3 = 3.9210$ Thus, $$ ext{Area}(R) = rac{1}{2} imes rac{ oot{2}}{4} (0 + 2(0.3240) + 2(1.3596) + 3.9210)$$ After computation, we find: $$ ext{Area}(R) ext{ } ightarrow 1.30$$Step 3
Use the substitution $u = x^2 + 2$ to show that the area of $R$ is $\frac{1}{2} \int^{\frac{ oot{2}}{2}}_{2} (u-2) \ln(u) \, du$.
Answer
To perform the substitution, we have:
ightarrow dx = \frac{du}{2x}$$ And from the substitution: 1. When $x = 0$, $u = 2$. 2. When $x = \frac{ oot{2}}{2}$, $u = 4$. Therefore, we rewriting the integral: $$ ext{Area}(R) = \int_{2}^{4} (u-2) \ln(u) \frac{du}{2}$$ Hence, $$ ext{Area}(R) = \frac{1}{2}\int_{2}^{4} (u-2) \ln(u) \, du$$Step 4
Hence, or otherwise, find the exact area of $R$.
Answer
To find the exact area, we need to compute the integral:
Using integration by parts, we let:
- ,
- , thus
Evaluating, we find:
\text{Area}(R) = \frac{1}{2}\Bigg [ igg( (u-2)\ln(u) - \int (u-2)\frac{1}{u}igg)\Bigg ]^{4}_{2}
After evaluating this expression between the bounds, we get:
- Substitute and .
The final answer gives:
This is the exact area of region .