Question

Let 𝐸 be the area of the region bounded by the curves 𝑦 = 𝑥² and 𝑦 = \(8\sqrt𝑥\), 𝑥≥0. Then 30𝐸 is equal to ________ (round off to 1 decimal place).

Answer 1

Correct Answer: 639

To determine the area \( E \) bounded by the curves \( y = x^2 \) and \( y = 8\sqrt{x} \) for \( x \geq 0 \), we first find their points of intersection. Set \( x^2 = 8\sqrt{x} \).
Rewrite it as \( x^4 = 64x \) or \( x^3 = 64 \), thus \( x = 4 \).
Therefore, the curves intersect at \( x = 0 \) and \( x = 4 \).
Now, integrate the difference between the functions from \( x=0 \) to \( x=4 \).
\( E = \int_0^4 (8\sqrt{x} - x^2) \, dx \). 
First solve \( \int (8\sqrt{x}) \, dx = \int (8x^{1/2}) \, dx = \frac{16}{3} x^{3/2} \).
Next solve \( \int (x^2) \, dx = \frac{x^3}{3} \).
Thus, \( E = \left[\frac{16}{3} x^{3/2}\right]_0^4 - \left[\frac{x^3}{3}\right]_0^4 \).
Calculate each term:
\(\frac{16}{3} (4)^{3/2} = \frac{16}{3} \times 8 = \frac{128}{3}\), \(\frac{4^3}{3} = \frac{64}{3}\).
\( E = \frac{128}{3} - \frac{64}{3} = \frac{64}{3}\).
To find \( 30E \), multiply: \( 30 \times \frac{64}{3} = 640 \).
Therefore, \( 30E = 640.0 \).
Verify that \( 640 \) is within the given range of \( 639, 639 \). As 640 lies slightly outside, ensure calculations align with context constraints.