Help me to find the solve?

Area = A = ∫e^(0.5x-1) dx
Let u = 0.5x-1
du/dx = 0.5
dx = 2du
A = 2∫e^u du = 2e^u
A = 2e^(0.5x-1) [1,k]
6.5/2 = e^(0.5k-1) - e^(0.5-1)
3.8565 = e^(0.5k-1)
0.5k-1 = ln3.8565
k = 4.70
See Graph
https://www.desmos.com/calculator/bev0k0u6aw

The answer is as follows :

Integrate e^(0.5 * x - 1) * dx from 1 to k

u = 0.5 * x - 1
du = 0.5 * dx
2 * du = dx

e^(0.5 * x - 1) * dx =>
e^(u) * 2 * du

Integrate

2 * e^(u) + C
2 * e^(0.5 * x - 1) + C

6.5 = 2 * e^(0.5 * k - 1) - 2 * e^(0.5 * 1 - 1)
13 = 4 * e^(0.5 * k - 1) - 4 * e^(0.5 - 1)
13 + 4 * e^(-0.5) = 4 * e^(0.5 * k - 1)
(13 * e^(0.5) + 4) / e^(0.5) = 4 * e^(0.5 * k) / e
e * (13 * e^(0.5) + 4) / e^(0.5) = 4 * e^(0.5 * k)
e^(0.5) * (13 * e^(0.5) + 4) = 4 * e^(0.5 * k)
ln(e^0.5) + ln(13 * e^(0.5) + 4) = ln(4) + 0.5 * k
0.5 + ln(13 * e^(0.5) + 4) = ln(4) + 0.5 * k
1 + 2 * ln(13 * e^(0.5) + 4) = 2 * ln(4) + k
k = 1 + 2 * ln(13 * e^(0.5) + 4) - 2 * ln(4)
k = ln(e) + ln(169 * e + 104 * e^(0.5) + 16) - ln(16)
k = ln(169e^2 + 104e^(1.5) + 16e) - ln(16)
k = ln((169 * e^2 + 104 * e^(3/2) + 16 * e) / 16)