Question

How do you find `\int _ { 0} ^ { 1} 3x ^ { 3} + 4x ^ { 2} + x - 2d x`?

Answer 1

`int_0^1 (3x^3+4x^2+x-2)dx=7/12`

Explanation:

As differential of `x^n` is `nx^(n-1)`, integral of `nx^(n-1)` is `x^n` and integral of `x^n` is `1/(n+1)x^(n+1)`.

Hence `int(3x^3+4x^2+x-2)dx`

= `3xxx^4/4+4xxx^3/3+x^2/2-2x+c`, where `c` is constant

= `(3x^4)/4+(4x^3)/3+x^2/2-2x+c`

and `int_0^1 (3x^3+4x^2+x-2)dx`

= `[(3x^4)/4+(4x^3)/3+x^2/2-2x+c]_0^1`

= `[(3xx1^4)/4+(4xx1^3)/3+(1xx1^2)/2-2xx1+c]-[(3xx0^4)/4+(4xx0^3)/3+(1xx0^2)/2-2xx0+c]`

= `3/4+4/3+1/2-2+c-0-0-0+0-c`

= `(9+16+6-24)/12`

= `7/12`