∫ln(9)0∫ln(5)0e10y−3xdydx=

Answer 1

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).