# What is the value of `int_6^8 t^2 - 18t + 4 dt`

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### 1 Answer

The value of `int_6^8 t^2 - 18t + 4 dt` has to be determined.

The integral `int x^n dx = x^(n+1)/(n+1)`

`int_6^8 t^2 - 18t + 4 dt`

=> `int_6^8 t^2 dt - int_6^8 18t dt + int_6^8 4 dt`

=> `(t^3/3)_6^8 - (18*t^2/2)_6^8 + (4t)_6^8`

=> `(t^3/3)_6^8 - (9*t^2)_6^8 + (4t)_6^8`

=> `(8^3 - 6^3)/3 - 9*(8^2 - 6^2) + 4(8 - 6)`

=> `296/3 - 252 + 8`

=> `-436/3`

**The definite integral `int_6^8 t^2 - 18t + 4 dt =-436/3` **