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Setler [38]
3 years ago
14

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.
Mathematics
1 answer:
Natalka [10]3 years ago
6 0

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})

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