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lorasvet [3.4K]
2 years ago
13

Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit

h the task of first evaluating sin(y2) dy. But it's impossible to do so in finite terms since sin(y2) dy is not an elementary function. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. Using this equation backward, we have 2 0 2 x sin(y2) dy dx
Mathematics
1 answer:
nignag [31]2 years ago
5 0

Answer:

Step-by-step explanation:

Given that:

\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\  \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\  where; \\ \\  D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}

\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}

\text{Now, above equation gives room for double integral  in  reverse order;}

\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA  \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy  \\ \\=  \int^2_0 ( y -0) \ sin (y^2) \ dy  \\ \\ = \int^2_0 y \ sin (y^2) \ dy  \\ \\  y^2 = U \\ \\  2y \ dy = du  \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du  \\ \\ = - \dfrac{1}{2} \Big [cos  \ U \Big]^2_o \\ \\ =  - \dfrac{1}{2} \Big [cos  \ (y^2)  \Big]^2_o  \\ \\ =  - \dfrac{1}{2} cos  (4) + \dfrac{1}{2} cos (0) \\ \\

=  - \dfrac{1}{2} cos  (4) + \dfrac{1}{2} (1) \\ \\  = \dfrac{1}{2}\Big [1- cos (4) \Big] \\ \\  = \mathbf{0.82682}

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