QUESTION 5 [TOTAL MARKS: 7 MARKS] An engineer analyzes the motion of a robot moving along a path where the velocity (in metres per seconds) at time $$ t $$ (in seconds) is given by $$ v(t)=t e^{t^{2}} $$. Using the substitution of $$ \boldsymbol{u}=\boldsymbol{t}^{2} $$, calculate the displacement from 0 second to 1 second. [7 marks]

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**Step 1:** Write the expression for displacement as an integral of the velocity:
$$
\text{Displacement} = \int_{0}^{1} v(t) \, dt = \int_{0}^{1} t e^{t^2} \, dt.
$$

**Step 2:** Use the substitution $$ u = t^2 $$. Compute the differential:
$$
\frac{du}{dt} = 2t \quad \Longrightarrow \quad du = 2t\, dt.
$$
Thus, we have:
$$
t\, dt = \frac{du}{2}.
$$

**Step 3:** Change the limits of integration:
- When $$ t = 0 $$, then $$ u = (0)^2 = 0 $$.
- When $$ t = 1 $$, then $$ u = (1)^2 = 1 $$.

**Step 4:** Rewrite the integral in terms of $$ u $$:
$$
\int_{0}^{1} t e^{t^2} \, dt = \int_{0}^{1} e^{u} \frac{du}{2} = \frac{1}{2} \int_{0}^{1} e^{u} \, du.
$$

**Step 5:** Evaluate the integral:
$$
\int_{0}^{1} e^{u} \, du = \left. e^{u} \right|_{0}^{1} = e^{1} - e^{0} = e - 1.
$$

**Step 6:** Multiply by the constant outside the integral:
$$
\text{Displacement} = \frac{1}{2} (e - 1).
$$
The displacement from 0 to 1 second is $$ \frac{1}{2}(e - 1) $$ meters.

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