Area between two curves

[canvas]

Calculate the area between two curves on interval [1,10]:

[math]y=\log_{10}(x)\,\,\,\,\,\wedge\,\,\,\,\,y=10^x\\\\A=\int_1^{10}(10^x-\log_{10}(x))dx=\left[\frac{10^x}{\ln(10)}-x\log_{10}(x)+\frac{x}{\ln(10)}\right]_1^{10}=...\\\\\text{Am I right? Or I'm completly stupid?}[/math]

 

[BigBeachBanana]

Looks right to me.
I would rewrite the integrand using the change of base, so it's a bit nicer. Just personal preference, but ultimately arrive at the same answer.
[math]10^x-\log_{10}x=10^x-\frac{\ln x}{\ln 10}[/math][math]\int 10^x-\frac{\ln x}{\ln 10}=\int10^x\,dx - \frac{1}{\ln 10}\int \ln x\,dx=\frac{-x\ln x + 10^x +x}{\ln(10)}+C[/math]

 

[canvas]

Looks right to me.
I would rewrite the integrand using the change of base, so it's a bit nicer. Just personal preference, but ultimately arrive at the same answer.
[math]10^x-\log_{10}x=10^x-\frac{\ln x}{\ln 10}[/math][math]\int 10^x-\frac{\ln x}{\ln 10}=\int10^x\,dx - \frac{1}{\ln 10}\int \ln x\,dx=\frac{-x\ln x + 10^x +x}{\ln(10)}+C[/math]

Thank you Sir, you are my savior!