Chapter 5: Q14P (page 247)

$\iint x^{2} e^{x^{2} y} d x d y$ over the area bounded by $y = x^{- x}, y = x^{- 2}$ and $x = I n 4$

Short Answer

Expert verified

The required solution is $4 - e I n 4$ .

Step by step solution

Definition of double integral

The double integral of $f (x, y)$ over the area Ain the $(x, y)$ plane as the limit of this sum, and we write it as $\iint_{A} f (x, y) d x d y$ .

Drawing the area bounded by the curve

The area bounded by the graph $y = \frac{1}{x}' y = \frac{1}{x_{2}}$ and $x = I n 4$ .

Intersecting points identification

From the figure the curves $\frac{1}{x}$ and $\frac{1}{x^{2}}$ intersect at $(x, y) = (1, 1)$ .

From the figure the curves and intersect at

Thus the integral is:

$\iint_{A} x^{2} e^{y x^{2}} d x d y = \int_{x = 1}^{I n 4} [\int_{y = 1 / x^{2}}^{1 / x} e^{y x^{2}} d (y x^{2})] d x = \int_{1}^{I n 4} (e^{x} - e) d x = {(e^{x} - x e)}_{1}^{I n 4} = 4 - e I n 4$

Therefore, the value is $4 - e I n 4$ .

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$\iint x^{2} e^{x^{2} y} d x d y$ over the area bounded by $y = x^{- x}, y = x^{- 2}$ and $x = I n 4$

Short Answer

Step by step solution

Definition of double integral

Drawing the area bounded by the curve

Intersecting points identification

From the figure the curves and intersect at

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∬x2ex2ydxdyover the area bounded byy=x-x,y=x-2andx=In4

Short Answer

Step by step solution

Definition of double integral

Drawing the area bounded by the curve

Intersecting points identification

From the figure the curves and intersect at

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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$\iint x^{2} e^{x^{2} y} d x d y$ over the area bounded by $y = x^{- x}, y = x^{- 2}$ and $x = I n 4$