dx dy is equal toa)4 log 8 + eb)8 log 8 - 16c)8 log 8 - 16 + ed)e - 16Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Answer

To evaluate the integral:

∫₁^{log 8} ∫₀^{log y} e^x+y dx dy,

we solve it by integrating with respect to x first, followed by y.

Step 1: Integrate with respect to x

The inner integral is:

∫₀^{log y} e^x+y dx.

Since e^x+y = e^x * e^y, we can rewrite this as:

= e^y ∫₀^{log y} e^x dx.

Integrating e^x with respect to x from 0 to log y:

∫₀^{log y} e^x dx = [ e^x ]₀^{log y} = y - 1.

Therefore, the inner integral becomes:

∫₀^{log y} e^x+y dx = e^y (y - 1).

Step 2: Integrate with respect to y

Now we substitute back into the outer integral:

∫₁^{log 8} e^y (y - 1) dy = ∫₁^{log 8} (y e^y - e^y) dy.

This expands to:

∫₁^{log 8} y e^y dy - ∫₁^{log 8} e^y dy.

Integral 1: ∫₁^{log 8} y e^y dy

To solve ∫ y e^y dy, we use integration by parts with:

- u = y and dv = e^y dy,
- then du = dy and v = e^y.

Using integration by parts:

∫ y e^y dy = y e^y - ∫ e^y dy = y e^y - e^y.

Evaluating this from 1 to log 8:

[ y e^y - e^y ]₁^{log 8} = ( (log 8) * 8 - 8 ) - ( e - e ) = 8 log 8 - 8.

Integral 2: ∫₁^{log 8} e^y dy

This integral is straightforward:

∫₁^{log 8} e^y dy = [ e^y ]₁^{log 8} = 8 - e.

Final Calculation

Putting it all together:

∫₁^{log 8} ∫₀^{log y} e^x+y dx dy = (8 log 8 - 8) - (8 - e).

Simplifying:

= 8 log 8 - 8 - 8 + e = 8 log 8 - 16 + e.

Conclusion:

The correct answer is: C: 8 log 8 - 16 + e.