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]

**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: \[ \text{Displacement} = \frac{1}{2} (e - 1). \]