04.( 2,0 pontos) Calcule o comprimento da curva \( \mathrm{f}(\mathrm{x})=x^{2} \) no intervalo \( 0 \leq \mathrm{x} \leq 2 \)

Para calcular o comprimento da curva \( f(x) = x^2 \) no intervalo \( 0 \leq x \leq 2 \), podemos usar a fórmula do comprimento de uma curva paramétrica. A fórmula do comprimento de uma curva paramétrica é dada por: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt \] Para a função \( f(x) = x^2 \), temos \( y = x^2 \). Portanto, \( \frac{dy}{dx} = 2x \). Agora, vamos calcular o comprimento da curva no intervalo \( 0 \leq x \leq 2 \) usando a fórmula acima. Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{0}^{2} \sqrt{\left(2x\right)^{2}+1} dx\) - step1: Rewrite the expression: \(\int_{0}^{2} \sqrt{4x^{2}+1} dx\) - step2: Evaluate the power: \(\int_{0}^{2} \left(4x^{2}+1\right)^{\frac{1}{2}} dx\) - step3: Evaluate the integral: \(\int \left(4x^{2}+1\right)^{\frac{1}{2}} dx\) - step4: Transform the expression: \(\int \left(\tan^{2}\left(t\right)+1\right)^{\frac{1}{2}}\times \frac{1}{2}\sec^{2}\left(t\right) dt\) - step5: Simplify the expression: \(\int \frac{1}{2}\left(\tan^{2}\left(t\right)+1\right)^{\frac{1}{2}}\sec^{2}\left(t\right) dt\) - step6: Simplify the expression: \(\int \frac{\sec^{3}\left(t\right)}{2} dt\) - step7: Simplify the expression: \(\int \frac{1}{2}\sec^{3}\left(t\right) dt\) - step8: Use properties of integrals: \(\frac{1}{2}\times \int \sec^{3}\left(t\right) dt\) - step9: Rewrite the expression: \(\frac{1}{2}\times \int \cos^{-3}\left(t\right) dt\) - step10: Evaluate the integral: \(\frac{1}{2}\left(\frac{\sec\left(t\right)\tan\left(t\right)}{2}+\frac{1}{2}\ln{\left(\left|\sec\left(t\right)+\tan\left(t\right)\right|\right)}\right)\) - step11: Calculate: \(\frac{\sec\left(t\right)\tan\left(t\right)}{4}+\frac{1}{4}\ln{\left(\left|\sec\left(t\right)+\tan\left(t\right)\right|\right)}\) - step12: Substitute back: \(\frac{\sec\left(\arctan\left(2x\right)\right)\tan\left(\arctan\left(2x\right)\right)}{4}+\frac{1}{4}\ln{\left(\left|\sec\left(\arctan\left(2x\right)\right)+\tan\left(\arctan\left(2x\right)\right)\right|\right)}\) - step13: Simplify: \(\frac{\left(1+4x^{2}\right)^{\frac{1}{2}}\times 2x}{4}+\frac{1}{4}\ln{\left(\left|\left(1+4x^{2}\right)^{\frac{1}{2}}+2x\right|\right)}\) - step14: Rewrite the expression: \(\frac{x\left(1+4x^{2}\right)^{\frac{1}{2}}}{2}+\frac{1}{4}\ln{\left(\left|\left(1+4x^{2}\right)^{\frac{1}{2}}+2x\right|\right)}\) - step15: Return the limits: \(\left(\frac{x\left(1+4x^{2}\right)^{\frac{1}{2}}}{2}+\frac{1}{4}\ln{\left(\left|\left(1+4x^{2}\right)^{\frac{1}{2}}+2x\right|\right)}\right)\bigg |_{0}^{2}\) - step16: Calculate the value: \(\frac{4\sqrt{17}+\ln{\left(\sqrt{17}+4\right)}}{4}\) O comprimento da curva \( f(x) = x^2 \) no intervalo \( 0 \leq x \leq 2 \) é \( \frac{4\sqrt{17}+\ln{(\sqrt{17}+4)}}{4} \).