Figure 1 shows part of the curve $y = \frac{3}{\sqrt{1+4x}}$. The region $R$ is bounded by the curve, the x-axis, and the lines $x = 0$ and $x = 2$, as shown shaded in Figure 1. (a) Use integration to find the area of $R$. The region $R$ is rotated $360^{\circ}$ about the x-axis. (b) Use integration to find the exact value of the volume of the solid formed.

A-Level Edexcel Maths Pure Vectors in 2 Dimensions: Figure 1 shows part of the curve

Figure 1 shows part of the curve $y = \frac{3}{\sqrt{1+4x}}$ - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 3

Question 4

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Figure 1 shows part of the curve $y = \frac{3}{\sqrt{1+4x}}$. The region $R$ is bounded by the curve, the x-axis, and the lines $x = 0$ and $x = 2$, as shown shaded ... show full transcript

Worked Solution & Example Answer:Figure 1 shows part of the curve $y = \frac{3}{\sqrt{1+4x}}$ - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 3

Step 1

Use integration to find the area of R.

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Answer

To find the area of the region $R$ , we will integrate the curve from $x = 0$ to $x = 2$ :

$A = \int_0^2 \frac{3}{\sqrt{1+4x}} \, dx$

We start with the substitution. Let:

$u = 1 + 4x$

Thus, when $x = 0$ , $u = 1$ and when $x = 2$ , $u = 9$ .

The differential $dx = \frac{1}{4} du$ .
Now substitute into the integral:

$A = \int_1^9 \frac{3}{\sqrt{u}} \cdot \frac{1}{4} \, du = \frac{3}{4} \int_1^9 u^{-\frac{1}{2}} \, du$
Evaluating the integral:

$\frac{3}{4} [2u^{\frac{1}{2}}]_{1}^{9} = \frac{3}{4} \left(2 \sqrt{9} - 2 \sqrt{1}\right)$

$= \frac{3}{4} (6 - 2) = \frac{3}{4} \cdot 4 = 3 \text{ units}^2$

Step 2

Use integration to find the exact value of the volume of the solid formed.

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Answer

To find the volume of the solid formed when the region $R$ is rotated about the x-axis, we will use the formula for the volume of revolution:

$V = \pi \int_0^2 \left(\frac{3}{\sqrt{1+4x}}\right)^2 \, dx$

Simplifying the expression:

$V = \pi \int_0^2 \frac{9}{1+4x} \, dx$
We will now apply the substitution method again:

Let $u = 1 + 4x$ , then $dx = \frac{1}{4} du$ . The limits change accordingly:

When $x = 0, u = 1$ and when $x = 2, u = 9$ .

Therefore:

$V = \pi \int_1^9 \frac{9}{u} \cdot \frac{1}{4} \, du = \frac{9\pi}{4} \int_1^9 u^{-1} \, du$
Evaluating this integral:

$V = \frac{9\pi}{4} [\ln|u|]_{1}^{9} = \frac{9\pi}{4} (\ln 9 - \ln 1) = \frac{9\pi}{4} \ln 9$

Finally, since $\\ln 1 = 0$ , we have:

$V = \frac{9\pi}{4} \ln 9 \text{ cubic units}$

Figure 1 shows part of the curve $y = \frac{3}{\sqrt{1+4x}}$ - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 3

Worked Solution & Example Answer:Figure 1 shows part of the curve $y = \frac{3}{\sqrt{1+4x}}$ - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 3

Use integration to find the area of R.

Use integration to find the exact value of the volume of the solid formed.

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