Question

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

Answer 1

Answer:

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

Step-by-step 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:

$\int {e^{2\cdot x^{3}}\cdot x^{2}} \, dx$

$\frac{1}{6}\int {e^{u}} \, du$

$f(u) = \frac{1}{6}\cdot e^{u} + C$

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

Where $C$ is the integration constant.

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

$\frac{1}{6}\cdot e^{2\cdot 0^{3}} + C= 0$

$C = -\frac{1}{6}$

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