Question

Please help me with the question; It is attached in the image

Answer 1

The function that passes through (0, 0) is
`f(x) = (1)/(6)\cdot e^{2\cdot x^(3)} - (1)/(6)`.

Explanation:

Firstly, we integrate the function by algebraic substitution:

`\int {x^(2)\cdot e^{2\cdot x^(3)}} \, dx` (1)

If
`u = 2\cdot x^(3)` and
`du = 6\cdot x^(2) dx`, then:

`\int {e^{2\cdot x^(3)}\cdot x^(2)} \, dx`

`(1)/(6)\int {e^(u)} \, du`

`f(u) = (1)/(6)\cdot e^(u) + C`

`f(x) = (1)/(6)\cdot e^{2\cdot x^(3)} + C`

Where
`C` is the integration constant.

If
`x = 0` and
`f(0) = 0`, then the integration constant is:

`(1)/(6)\cdot e^{2\cdot 0^(3)} + C= 0`

`C = -(1)/(6)`

Hence, the function that passes through (0, 0) is
`f(x) = (1)/(6)\cdot e^{2\cdot x^(3)} - (1)/(6)`.