# Calculate the definite integral of f(x)=5*(8^x)+x^2 from x=-1 to x=0

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### 1 Answer

f(x)= 5(8^x) + x^2 from x = 01 to 0.

`==gt int_-1^0 5(8^x) + x^2 dx = 5int_-1^0 8^x dx + int_-1^0 x^2 dx`

`` First we will find the intergral of 8^x

==> We know that e^ln 8 = 8

==> `int_-1^0 8^x dx = int_-1^0 (e^ln8)^x dx`

`= int_-1^0 e^(xln8) dx `

`` Let u= xln8 ==> du = ln8 dx

`==> int_-1^0 e^u (du)/ln8 = (1/ln8) int_-1^0 e^u du `

`= (1/ln8) e^u= e^u/ln8 `

`=( e^xln8) /ln8= (e^ln8)^x /ln8= 8^x/ln8`

`==> int 8^x dx = 8^x/ln8 `

`==> 5int_-1^0 8^x dx = 5(8^0/ln8 - 8^-1 /ln8)`

`= 5(1/ln8 - 1/(8ln8)) `

`= 5(7/(8ln8)) = 35/(8ln8)`

`Now we will find the integral of x^2 `

`==> int_-1^0 x^2 dx = x^3 /3 `

`= (0^3 /3 - (-1^3)/3 = 1/3`

`==> int_-1^0 5(8^x) + x^2 dx = 35/(8ln8) + 1/3`

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