∫ln(9)0∫ln(5)0e10y−3xdydx=
To evaluate the given integral, we can first integrate with respect to y and then with respect to x.
∫ln(9)0∫ln(5)0e^(10y−3x)dydx
Let's start with the inner integral with respect to y.
∫ln(9)0 e^(10y−3x)dy
To integrate e^(10y-3x) with respect to y, we treat x as a constant.
∫ln(9)0 e^(10y) e^(-3x) dy
Using the property that ∫e^(ax) dx = (1/a)e^(ax), we can integrate e^(10y) with respect to y.
(1/10)e^(10y) e^(-3x) from y=0 to y=ln(9)
= (1/10)(e^(10ln(9)-3x) - e^(0-3x))
= (1/10)(e^(ln(9^10)-3x) - e^(-3x))
= (1/10)(e^(ln(9^10/1) -3x) - e^(-3x))
= (1/10)(e^(ln(9^10))e^(-3x) - e^(-3x))
= (1/10)((9^10)e^(-3x) - e^(-3x))
= (1/10)(9^10 - 1)e^(-3x)
Now, we can integrate the resulting expression with respect to x.
∫ln(5)0 (1/10)(9^10 - 1)e^(-3x) dx
Using the property that ∫e^(ax) dx = (1/a)e^(ax), we integrate e^(-3x) with respect to x.
(1/10)(9^10 - 1)(1/-3)e^(-3x) from x=0 to x=ln(5)
= (1/10)(9^10 - 1)(1/-3)(e^(-3ln(5)) - e^(-3(0)))
= (1/10)(9^10 - 1)(1/-3)((1/5^3) - 1)
= (1/10)(9^10 - 1)(-1/3)((1/125) - 1)
Simplifying further, we get:
= -(1/300)(9^10 - 1)(1 - 125)
= -(1/300)(9^10 - 1)(-124)
= (124/300)(9^10 - 1)
Therefore, ∫ln(9)0∫ln(5)0e^(10y−3x)dydx = (124/300)(9^10 - 1).
To evaluate the double integral ∫∫e^(10y - 3x) dydx over the region specified, we need to follow these steps:
Step 1: Integrate with respect to y:
∫(ln(9) 0) e^(10y - 3x) dy
To integrate e^(10y - 3x) with respect to y, we treat x as a constant:
= (1/10) e^(10y - 3x) + C1
Step 2: Evaluate the above expression at the limits of integration:
= (1/10) e^(10(ln(9)) - 3x) + C1
- (1/10) e^(-3x) + C1
Step 3: Integrate with respect to x:
∫(ln(5) 0) [(1/10) e^(10(ln(9)) - 3x) + C1] dx
To integrate [(1/10) e^(10(ln(9)) - 3x) + C1] with respect to x, we treat C1 as a constant:
= [(1/10)(e^(10(ln(9)) - 3x) / -3) + C1x] + C2
Step 4: Evaluate the above expression at the limits of integration:
= [(1/10)(e^(10(ln(9)) - 3(0)) / -3) + C1(0)] - [(1/10)(e^(10(ln(9)) - 3(ln(5)))) / -3) + C1(ln(5))] + C2
Simplifying further:
= [(1/10)(e^(10ln(9)) / -3) + 0] - [(1/10)(e^(10ln(9)) * e^(-3ln(5)))) / -3) + 0] + C2
= [(1/10)(9^10) / -3] - [(1/10)(9^10 * 5^(-3))] + C2
Finally, the result of the double integral is:
[(9^10) / (-30)] - [(9^10) / (50)] + C2