Vamos resolver passo a passo a integral de linha: \[ \int_C (x\,dx + y\,dy + z\,dz) \] com \( C \) definida pela curva paramétrica: \[ \gamma(t) = (2t^2, t^3, t), \quad 0 \leq t \leq 1. \] --- Passo 1: Expressar \(x, y, z\) e suas derivadas em função de \(t\): \[ x = 2t^2 \implies dx = \frac{d}{dt}(2t^2) dt = 4t\, dt \] \[ y = t^3 \implies dy = \frac{d}{dt}(t^3) dt = 3t^2\, dt \] \[ z = t \implies dz = \frac{d}{dt}(t) dt = dt \] --- Passo 2: Substituir na integral: \[ \int_0^1 \left[ x \frac{dx}{dt} + y \frac{dy}{dt} + z \frac{dz}{dt} \right] dt = \int_0^1 \left[ x\,dx + y\,dy + z\,dz \right] \] Substituindo: \[ = \int_0^1 \left[ (2t^2)(4t) + (t^3)(3t^2) + (t)(1) \right] dt \] \[ = \int_0^1 \left[ 8t^3 + 3t^5 + t \right] dt \] --- Passo 3: Calcular a integral: \[ \int_0^1 8t^3 dt = 8 \cdot \frac{t^4}{4} \Big|_0^1 = 8 \cdot \frac{1}{4} = 2 \] \[ \int_0^1 3t^5 dt = 3 \cdot \frac{t^6}{6} \Big|_0^1 = 3 \cdot \frac{1}{6} = \frac{1}{2} \] \[ \int_0^1 t dt = \frac{t^2}{2} \Big|_0^1 = \frac{1}{2} \] Somando tudo: \[ 2 + \frac{1}{2} + \frac{1}{2} = 2 + 1 = 3 \] --- Resposta correta: B) \(3\)