Question

Evaluate the integral Integral from 0 to 1 Integral from 0 to 3 Integral from 3 y to 9 StartFraction 6 cosine (x squared )Over 5 StartRoot z EndRoot EndFraction dx dy dz by changing the order of integration in an appropriate way.

Answer 1

Answer:

$\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz$ $=\frac{18}{5}(1+\frac{sin2}{2})$

Step-by-step explanation:

cosine x²= cos x²

Rule

• $\int x^ndx= \frac{x^{n+1}}{n+1}+c$

• $\int cos \ mx \ dx = \frac{sin \ mx}{m}+c$
• $\int \frac{1}{\sqrt x}dx = \frac{\sqrt x}{\frac{1}{2}} +c= 2\sqrt x+c$

Given that,

$\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz$

$=\int ^1_0[\int^3_0(\int^9_{3y} \frac{6cos x^2}{5\sqrt z}dz)dy]dz$

$=\int^1_0[\int^3_0([\frac{6cos x^2 \times \sqrt z}{5\times \frac{1}{2}}]^9_{3y})dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( \sqrt 9-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( 3-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\frac{12cos x^2 \times( 3y-\frac{\sqrt{3}y^\frac{3}{2}}{\frac{3}{2}})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 3.3-\frac{2\sqrt{3}.3^\frac{3}{2}}{3})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 9-6)}{5}]dx$

$=\frac{18}{5}\int^1_02cos x^2dx$

$=\frac{18}{5}\int^1_0(1+cos2x)dx$

$=\frac{18}{5}[(x+\frac{sin2x}{2})]^1_0$

$=\frac{18}{5}(1+\frac{sin2}{2})$