The area bounded by the curve y = x sin ( 2 x ) and the X-axis between the lines x = 0 and x = 4 π (in square units), is....

Question

The area bounded by the curve y = x ​ sin ( 2 x ​ ) ​ and the X-axis between the lines x = 0 and x = 4 π (in square units), is....

TestHub · Accepted Answer

The required area is given by the integral:

$A = \int_{0}^{4 π} x sin (\frac{x}{2}) d x$

Since $sin (\frac{x}{2}) \geq 0$ for $x \in [0, 2 π]$ and $sin (\frac{x}{2}) \leq 0$ for $x \in [2 π, 4 π]$ , the integral can be split as:

$A = \int_{0}^{2 π} x sin (\frac{x}{2}) d x - \int_{2 π}^{4 π} x sin (\frac{x}{2}) d x$

Using integration by parts:

$\int x sin (\frac{x}{2}) d x = x (- 2 cos (\frac{x}{2})) - \int 1 \cdot (- 2 cos (\frac{x}{2})) d x$

$= - 2 x cos (\frac{x}{2}) + 4 sin (\frac{x}{2})$

Evaluating the first integral:

$\int_{0}^{2 π} x sin (\frac{x}{2}) d x = [- 2 x cos (\frac{x}{2}) + 4 sin (\frac{x}{2})]_{0}^{2 π} = (- 4 π (- 1) + 0) - (0 + 0) = 4 π$

Evaluating the second integral:

$\int_{2 π}^{4 π} x sin (\frac{x}{2}) d x = [- 2 x cos (\frac{x}{2}) + 4 sin (\frac{x}{2})]_{2 π}^{4 π} = (- 8 π (1) + 0) - (- 4 π (- 1) + 0) = - 8 π - 4 π = - 12 π$

Substituting these values back into the area expression:

$A = 4 π - (- 12 π) = 16 π$

Answer: $16 π$

Mathematics - Area Under Curves Question with Solution | TestHub

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