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Estimating and Calculating Areas Under Curves

This lesson explores the region bounded by the curve y = sqrt(x) ln(2x) from x = 1 to 4, demonstrating how to approximate area using the trapezium rule and find the exact area using integration by parts.

Scene 1 of 4
Visualizing the Region
R14
y=xln⁡(2x)y = \sqrt{x} \ln(2x)
Look at the shaded region R. It is trapped between our curve, the x-axis, and two vertical lines at x equals 1 and 4.
Step-by-step solver
1

Trapezium Rule setup

With 3 strips from 1 to 4, width h = 1. Coordinates are x_0=1, x_1=2, x_2=3, x_3=4. Calculate y_n = sqrt(x_n) ln(2x_n).

y0≈0.693,y1≈2.443,y2≈4.075,y3≈5.773y_0 \approx 0.693, y_1 \approx 2.443, y_2 \approx 4.075, y_3 \approx 5.773
2

Sum the trapeziums

Area = (h/2) * [y_0 + 2(y_1 + y_2) + y_3].

A≈0.5∗[0.693+2(2.443+4.075)+5.773]≈9.75A \approx 0.5 * [0.693 + 2(2.443 + 4.075) + 5.773] \approx 9.75
3

Integration by parts

Let u = ln(2x), dv = x^(1/2) dx. Then du = 1/x dx, v = (2/3)x^(3/2).

∫udv=uv−∫vdu=23x3/2ln⁡(2x)−23∫x1/2dx\int u dv = uv - \int v du = \frac{2}{3}x^{3/2}\ln(2x) - \frac{2}{3}\int x^{1/2} dx
4

Evaluate integral

The result is (2/3)x^(3/2)ln(2x) - (4/9)x^(3/2) + C.

∫14=[23x3/2ln⁡(2x)−49x3/2]14\int_1^4 = [\frac{2}{3}x^{3/2}\ln(2x) - \frac{4}{9}x^{3/2}]_1^4
5

Exact Result

Substitute 4 and 1 to find the final value in the requested form.

(163ln⁡8−329)−(23ln⁡2−49)=15ln⁡2−289(\frac{16}{3}\ln 8 - \frac{32}{9}) - (\frac{2}{3}\ln 2 - \frac{4}{9}) = 15\ln 2 - \frac{28}{9}

Original question

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