ISI2015-MMA-77

Question

Let $R$ be the triangle in the $xy$ – plane bounded by the $x$-axis, the line $y=x$, and the line $x=1$. The value of the double integral $$ \int \int_R \frac{\sin x}{x}\: dxdy$$ is

• A. $1-\cos 1$
• B. $\cos 1$
• C. $\frac{\pi}{2}$
• D. $\pi$

Comments

http://www.stankova.net/statistics_2012/double_integration.pdf

Answer 1

$$\int\int_D\frac{\sin x}{x}dx\ dy$$ We cannot integrate $\frac{\sin x}{x}$ directly hence we first try to integrate dy first SInce when we draw vertical line through D ,its enters D at y = 0 and leaves at y = x,hence $$=\int_{0}^{1}\int_{0}^{x}\frac{\sin x}{x}dy\ dx$$ $$=\int_{0}^{1}[\frac{\sin x}{x} * y]_0^{x}\ dx$$ $$=\int_{0}^{1}{\sin x}\ dx$$ $$=[-\cos x]_{0}^{1}$$ $$=\cos 0 - \cos1$$ $$=1 - \cos1$$