Answer:
The function that passes through (0, 0) is [tex]f(x) = \frac{1}{6}\cdot e^{2\cdot x^{3}} - \frac{1}{6}[/tex].
Step-by-step explanation:
Firstly, we integrate the function by algebraic substitution:
[tex]\int {x^{2}\cdot e^{2\cdot x^{3}}} \, dx[/tex] (1)
If [tex]u = 2\cdot x^{3}[/tex] and [tex]du = 6\cdot x^{2} dx[/tex], then:
[tex]\int {e^{2\cdot x^{3}}\cdot x^{2}} \, dx[/tex]
[tex]\frac{1}{6}\int {e^{u}} \, du[/tex]
[tex]f(u) = \frac{1}{6}\cdot e^{u} + C[/tex]
[tex]f(x) = \frac{1}{6}\cdot e^{2\cdot x^{3}} + C[/tex]
Where [tex]C[/tex] is the integration constant.
If [tex]x = 0[/tex] and [tex]f(0) = 0[/tex], then the integration constant is:
[tex]\frac{1}{6}\cdot e^{2\cdot 0^{3}} + C= 0[/tex]
[tex]C = -\frac{1}{6}[/tex]
Hence, the function that passes through (0, 0) is [tex]f(x) = \frac{1}{6}\cdot e^{2\cdot x^{3}} - \frac{1}{6}[/tex].
