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Figure 3 shows a sketch of the curve with equation $y = \sqrt{x}$ - Edexcel - A-Level Maths Pure - Question 7 - 2019 - Paper 2

Question 7

$Figure-3-shows-a-sketch-of-the-curve-with-equation--$y-=-\sqrt{x}$-Edexcel-A-Level Maths Pure-Question 7-2019-Paper 2.png$

Figure 3 shows a sketch of the curve with equation $y = \sqrt{x}$. The point $P(x, y)$ lies on the curve. The rectangle, shown shaded on Figure 3, has height $y$ an... show full transcript

Worked Solution & Example Answer:Figure 3 shows a sketch of the curve with equation $y = \sqrt{x}$ - Edexcel - A-Level Maths Pure - Question 7 - 2019 - Paper 2

Step 1

States $\int \sqrt{x} \,dx$ with or without the 'dx'

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Answer

$\int \sqrt{x} \,dx$ is given by $\frac{2}{3}x^{3/2} + C$ .

Step 2

Integrates $\sqrt{x}$ to give $\int_{4}^{9} \sqrt{x} \,dx$

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Answer

To compute the definite integral, we evaluate: $\int_{4}^{9} \sqrt{x} \,dx = \left[ \frac{2}{3}x^{3/2} \right]_{4}^{9} = \frac{2}{3}(9^{3/2}) - \frac{2}{3}(4^{3/2})$

Calculating each part:

$9^{3/2} = 27$
$4^{3/2} = 8$

Thus, $\int_{4}^{9} \sqrt{x} \,dx = \frac{2}{3}(27) - \frac{2}{3}(8) = \frac{54}{3} - \frac{16}{3} = \frac{38}{3}$ .

Step 3

Final Calculation

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Answer

The limit can now be computed as follows: $\lim_{\delta x \to 0} \sum_{x=4}^{9} \sqrt{x} \,\delta x = \int_{4}^{9} \sqrt{x} \,dx = \frac{38}{3},$ or approximately 12.7.

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