Answer :
To solve the integral [tex]\(\int_0^1 \int_{\sqrt{x}}^1 y^3 e^{x y} \, dy \, dx\)[/tex] by changing the order of integration, follow these steps:
1. Understand the original region of integration:
The original bounds for the integral [tex]\(\int_0^1 \int_{\sqrt{x}}^1 y^3 e^{x y} \, dy \, dx\)[/tex] are:
- [tex]\(0 \leq x \leq 1\)[/tex]
- [tex]\(\sqrt{x} \leq y \leq 1\)[/tex]
2. Sketch the region:
To change the order of integration, it is helpful to sketch the region in the [tex]\(xy\)[/tex]-plane.
- Since [tex]\(0 \leq x \leq 1\)[/tex] and [tex]\(\sqrt{x} \leq y \leq 1\)[/tex], start by plotting [tex]\(y = \sqrt{x}\)[/tex].
- Note that [tex]\(y = 1\)[/tex] is the upper bound. The curve [tex]\(y = \sqrt{x}\)[/tex] intersects [tex]\(y = 1\)[/tex] when [tex]\(x = 1\)[/tex].
- Therefore, the region of integration is bounded by [tex]\(y = \sqrt{x}\)[/tex], [tex]\(y = 1\)[/tex], [tex]\(x = 0\)[/tex], and [tex]\(x = 1\)[/tex].
3. Determine new bounds for integration:
Changing the order of integration involves rewriting the region in terms of [tex]\(y\)[/tex] first and then [tex]\(x\)[/tex].
- The [tex]\(y\)[/tex]-values range from [tex]\(0\)[/tex] to [tex]\(1\)[/tex].
- For a fixed [tex]\(y\)[/tex], [tex]\(x\)[/tex] ranges from [tex]\(0\)[/tex] to [tex]\(y^2\)[/tex] (since [tex]\(y = \sqrt{x} \Rightarrow x = y^2\)[/tex]).
So, the new bounds are:
- [tex]\(0 \leq y \leq 1\)[/tex]
- [tex]\(0 \leq x \leq y^2\)[/tex]
4. Rewriting the integral with new bounds:
The integral can be expressed as:
[tex]\[ \int_0^1 \int_0^{y^2} y^3 e^{x y} \, dx \, dy \][/tex]
5. Evaluate the inner integral:
The inner integral is with respect to [tex]\(x\)[/tex]:
[tex]\[ \int_0^{y^2} y^3 e^{x y} \, dx \][/tex]
Treat [tex]\(y\)[/tex] as a constant:
[tex]\[ y^3 \int_0^{y^2} e^{x y} \, dx \][/tex]
The antiderivative of [tex]\(e^{x y}\)[/tex] with respect to [tex]\(x\)[/tex] is [tex]\(\frac{e^{x y}}{y}\)[/tex]. Evaluate from [tex]\(0\)[/tex] to [tex]\(y^2\)[/tex]:
[tex]\[ y^3 \left. \frac{e^{x y}}{y} \right|_0^{y^2} = y^3 \left( \frac{e^{y y^2}}{y} - \frac{e^0}{y} \right) \][/tex]
Simplify:
[tex]\[ y^3 \left( \frac{e^{y^3}}{y} - \frac{1}{y} \right) = y^3 \left( \frac{e^{y^3} - 1}{y} \right) = y^2 (e^{y^3} - 1) \][/tex]
6. Evaluate the outer integral:
The outer integral is:
[tex]\[ \int_0^1 y^2 (e^{y^3} - 1) \, dy \][/tex]
Break it into two integrals:
[tex]\[ \int_0^1 y^2 e^{y^3} \, dy - \int_0^1 y^2 \, dy \][/tex]
For the first integral, use the substitution [tex]\(u = y^3 \Rightarrow du = 3y^2 \, dy \Rightarrow dy = \frac{du}{3y^2}\)[/tex]:
[tex]\[ \int_0^1 y^2 e^{y^3} \, dy = \int_0^1 e^u \frac{du}{3y^2} = \int_0^1 \frac{e^u}{3y^2} \cdot \frac{du}{3y^2} = \int_0^1 \frac{e^u}{3} \, du = \frac{1}{3} \int_0^1 e^u \, du = \frac{1}{3} \left[ e^u \right]_0^1 = \frac{1}{3} (e - 1) \][/tex]
For the second integral:
[tex]\[ \int_0^1 y^2 \, dy = \left. \frac{y^3}{3} \right|_0^1 = \frac{1}{3} \][/tex]
7. Combine the results:
So, the original integral becomes:
[tex]\[ \frac{1}{3} (e - 1) - \frac{1}{3} = \frac{e - 2}{3} \][/tex]
Thus, the value of the integral is:
[tex]\[ \boxed{-\frac{2}{3} + \frac{E}{3}} \][/tex]
1. Understand the original region of integration:
The original bounds for the integral [tex]\(\int_0^1 \int_{\sqrt{x}}^1 y^3 e^{x y} \, dy \, dx\)[/tex] are:
- [tex]\(0 \leq x \leq 1\)[/tex]
- [tex]\(\sqrt{x} \leq y \leq 1\)[/tex]
2. Sketch the region:
To change the order of integration, it is helpful to sketch the region in the [tex]\(xy\)[/tex]-plane.
- Since [tex]\(0 \leq x \leq 1\)[/tex] and [tex]\(\sqrt{x} \leq y \leq 1\)[/tex], start by plotting [tex]\(y = \sqrt{x}\)[/tex].
- Note that [tex]\(y = 1\)[/tex] is the upper bound. The curve [tex]\(y = \sqrt{x}\)[/tex] intersects [tex]\(y = 1\)[/tex] when [tex]\(x = 1\)[/tex].
- Therefore, the region of integration is bounded by [tex]\(y = \sqrt{x}\)[/tex], [tex]\(y = 1\)[/tex], [tex]\(x = 0\)[/tex], and [tex]\(x = 1\)[/tex].
3. Determine new bounds for integration:
Changing the order of integration involves rewriting the region in terms of [tex]\(y\)[/tex] first and then [tex]\(x\)[/tex].
- The [tex]\(y\)[/tex]-values range from [tex]\(0\)[/tex] to [tex]\(1\)[/tex].
- For a fixed [tex]\(y\)[/tex], [tex]\(x\)[/tex] ranges from [tex]\(0\)[/tex] to [tex]\(y^2\)[/tex] (since [tex]\(y = \sqrt{x} \Rightarrow x = y^2\)[/tex]).
So, the new bounds are:
- [tex]\(0 \leq y \leq 1\)[/tex]
- [tex]\(0 \leq x \leq y^2\)[/tex]
4. Rewriting the integral with new bounds:
The integral can be expressed as:
[tex]\[ \int_0^1 \int_0^{y^2} y^3 e^{x y} \, dx \, dy \][/tex]
5. Evaluate the inner integral:
The inner integral is with respect to [tex]\(x\)[/tex]:
[tex]\[ \int_0^{y^2} y^3 e^{x y} \, dx \][/tex]
Treat [tex]\(y\)[/tex] as a constant:
[tex]\[ y^3 \int_0^{y^2} e^{x y} \, dx \][/tex]
The antiderivative of [tex]\(e^{x y}\)[/tex] with respect to [tex]\(x\)[/tex] is [tex]\(\frac{e^{x y}}{y}\)[/tex]. Evaluate from [tex]\(0\)[/tex] to [tex]\(y^2\)[/tex]:
[tex]\[ y^3 \left. \frac{e^{x y}}{y} \right|_0^{y^2} = y^3 \left( \frac{e^{y y^2}}{y} - \frac{e^0}{y} \right) \][/tex]
Simplify:
[tex]\[ y^3 \left( \frac{e^{y^3}}{y} - \frac{1}{y} \right) = y^3 \left( \frac{e^{y^3} - 1}{y} \right) = y^2 (e^{y^3} - 1) \][/tex]
6. Evaluate the outer integral:
The outer integral is:
[tex]\[ \int_0^1 y^2 (e^{y^3} - 1) \, dy \][/tex]
Break it into two integrals:
[tex]\[ \int_0^1 y^2 e^{y^3} \, dy - \int_0^1 y^2 \, dy \][/tex]
For the first integral, use the substitution [tex]\(u = y^3 \Rightarrow du = 3y^2 \, dy \Rightarrow dy = \frac{du}{3y^2}\)[/tex]:
[tex]\[ \int_0^1 y^2 e^{y^3} \, dy = \int_0^1 e^u \frac{du}{3y^2} = \int_0^1 \frac{e^u}{3y^2} \cdot \frac{du}{3y^2} = \int_0^1 \frac{e^u}{3} \, du = \frac{1}{3} \int_0^1 e^u \, du = \frac{1}{3} \left[ e^u \right]_0^1 = \frac{1}{3} (e - 1) \][/tex]
For the second integral:
[tex]\[ \int_0^1 y^2 \, dy = \left. \frac{y^3}{3} \right|_0^1 = \frac{1}{3} \][/tex]
7. Combine the results:
So, the original integral becomes:
[tex]\[ \frac{1}{3} (e - 1) - \frac{1}{3} = \frac{e - 2}{3} \][/tex]
Thus, the value of the integral is:
[tex]\[ \boxed{-\frac{2}{3} + \frac{E}{3}} \][/tex]
