# What is the definite integral of 4x^3 + 3x^2 + 1/x between x = 2 and x = 6

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justaguide | Certified Educator

The definite integral of `4x^3 + 3x^2 + 1/x` between x = 2 and x = 6 has to be determined.

The indefinite integral of` 4x^3 + 3x^2 + 1/x` is `x^4 + x^3 + ln x`

`int_(2)^6 4x^3 + 3x^2 + 1/x dx`

=> `6^4 - 2^4 + 6^3 - 2^3 + ln 6 - ln 2`

=> `1296 - 16 + 216 - 8 + ln 3`

=> `1488 + ln 3`

`~~ 1489.09`

**The required definite integral is 1488 + ln 3**