solve this---**Question Stem:**
The given integral expression is:
$$ \int_{1}^{e} \int_{0}^{\ln x} f(x, y) dy dx $$
**Options:**
1) $$ \int_{0}^{1} \int_{0}^{\ln y} f(x, y) dx dy $$
2) $$ \int_{0}^{1} \int_{\ln y}^{0} f(x, y) dx dy $$
3) $$ \int_{0}^{1} \int_{e}^{x} f(x, y) dx dy $$ (Note: The lower limit of the inner integral appears to be `e^y` based on common mathematical notation context and option 4, but it is written as `ex`. Assuming it might be a typo and intended as `e^y` or `e^x` - if `e^x` it makes less sense for `dx dy` order. Given the image, it strictly looks like `ex` but this is likely an optical illusion or bad rendering for `e^x` or `e^y`. Given the options, it most likely means `e^y`. For precise extraction, I will write `ex` as it appears.)
4) $$ \int_{0}^{1} \int_{e}^{y} f(x, y) dx dy $$ (Note: The lower limit of the inner integral appears to be `e^y`.)
**Correction/Clarification on Option 3 and 4:**
Upon closer inspection and considering the context of changing the order of integration for the given problem, the lower limit of the inner integral in option 3 is written as `e` but the variable in the exponent is `x`, making it `e^x`. The lower limit of the inner integral in option 4 is written as `e` but the variable in the exponent is `y`, making it `e^y`.
Let's re-extract them clearly assuming standard mathematical notation for exponents:
**Corrected Options (interpreting exponents based on typical math notation):**
1) $$ \int_{0}^{1} \int_{0}^{\ln y} f(x, y) dx dy $$
2) $$ \int_{0}^{1} \int_{\ln y}^{0} f(x, y) dx dy $$
3) $$ \int_{0}^{1} \int_{e^x}^{e} f(x, y) dx dy $$ (Based on visual `e` and `x` next to it, likely `e^x`)
4) $$ \int_{0}^{1} \int_{e^y}^{e} f(x, y) dx dy $$ (Based on visual `e` and `y` next to it, likely `e^y`)
**Strict Literal Extraction (as characters appear):**
1) $$ \int_{0}^{1} \int_{0}^{\ln y} f(x, y) dx dy $$
2) $$ \int_{0}^{1} \int_{\ln y}^{0} f(x, y) dx dy $$
3) $$ \int_{0}^{1} \int_{e x}^{e} f(x, y) dx dy $$ (The characters `ex` appear as `e` followed by `x` at the lower limit)
4) $$ \int_{0}^{1} \int_{e y}^{e} f(x, y) dx dy $$ (The characters `ey` appear as `e` followed by `y` at the lower limit)
For the purpose of accuracy, I will provide the "Strict Literal Extraction" as it directly reflects what is presented, while acknowledging the common interpretation for mathematical expressions.
**Final Extraction (Strict Literal):**
**Question Stem:**
The integral expression is:
`Integral from 1 to e of (Integral from 0 to ln(x) of f(x, y) dy) dx`
(Represented as: `∫[1,e] ∫[0,ln(x)] f(x, y) dy dx`)
**Options:**
1) `Integral from 0 to 1 of (Integral from 0 to ln(y) of f(x, y) dx) dy`
(Represented as: `∫[0,1] ∫[0,ln(y)] f(x, y) dx dy`)
2) `Integral from 0 to 1 of (Integral from ln(y) to 0 of f(x, y) dx) dy`
(Represented as: `∫[0,1] ∫[ln(y),0] f(x, y) dx dy`)
3) `Integral from 0 to 1 of (Integral from e x to e of f(x, y) dx) dy`
(Represented as: `∫[0,1] ∫[ex,e] f(x, y) dx dy`)
4) `Integral from 0 to 1 of (Integral from e y to e of f(x, y) dx) dy`
(Represented as: `∫[0,1] ∫[ey,e] f(x, y) dx dy`)
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Let's solve this double integral problem by changing the order of integration. We start with the integral from 1 to e, and from 0 to natural log of x, of f of x y, dy dx. This represents integrating over a specific region in the xy-plane. The outer integral has x ranging from 1 to e, while the inner integral has y ranging from 0 to ln x for each value of x.
Now let's sketch the region of integration. The bounds tell us that x ranges from 1 to e, and for each x, y ranges from 0 to ln x. The boundary curve is y equals ln x. We draw vertical lines at x equals 1 and x equals e, and the curve y equals ln x. The region is bounded by these curves, with key points at (1,0) and (e,1). This shaded region represents our domain of integration.
To change the order of integration, we need to express the same region with y as the outer variable and x as the inner variable. From the boundary curve y equals ln x, we solve for x to get x equals e to the y. Now y ranges from 0 to 1, and for each y, x ranges from e to the y to e. We integrate horizontally across the region instead of vertically, but we're covering the exact same area.
Now we construct the new integral with reversed order. The original integral from 1 to e, from 0 to ln x, becomes the integral from 0 to 1, from e to the y to e, of f x y, dx dy. The blue lines show the original vertical integration strips, while the orange lines show the new horizontal integration strips. Both methods cover the exact same region, just with different integration paths.
Let's evaluate each option systematically. Option 1 has wrong bounds with ln y instead of e to the y. Option 2 has the limits reversed, giving negative integration. Option 3 uses ex instead of e to the y, which is incorrect notation. Option 4 gives us the integral from 0 to 1, from e to the y to e, which matches our derived bounds perfectly. Therefore, the correct answer is option 4.