Chapter 5: Q32P (page 248)

$\int_{y = 0}^{1} \int_{x = y^{2}}^{1} \frac{e^{x}}{\sqrt{x}} dxdy$

Short Answer

Expert verified

The required solution is $e - 1$ .

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 $\int \int_{A} f (x, y) dxdy .$

Drawing the area bounded by the curve

The area of integration or bounded region is shown below,

Calculation of the area under the curve changing boundaries

Changing the boundaries of the given integration:

$x : 0 \to 1, and y : 0 \to \sqrt{x}$

So, the integration

$l = \int_{0}^{1} dx \frac{e^{x}}{\sqrt{x}} \int_{0}^{\sqrt{x}} dy = \int_{0}^{1} e^{x} dx = {e^{x}}_{0}^{1} = e - 1$

Therefore, the value of the integration is, $e - 1 .$

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$\int_{y = 0}^{1} \int_{x = y^{2}}^{1} \frac{e^{x}}{\sqrt{x}} dxdy$

Short Answer

Step by step solution

Definition of double integral

Drawing the area bounded by the curve

Calculation of the area under the curve changing boundaries

One App. One Place for Learning.

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∫y=01∫x=y21exxdxdy

Short Answer

Step by step solution

Definition of double integral

Drawing the area bounded by the curve

Calculation of the area under the curve changing boundaries

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Magnetism

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Study anywhere. Anytime. Across all devices.

$\int_{y = 0}^{1} \int_{x = y^{2}}^{1} \frac{e^{x}}{\sqrt{x}} dxdy$