find the arclength of $y = 3x^{2}-5$ on $0leq xleq1$

Explanation:

Step1: Find the derivative of y

The derivative of $y = 3x^{2}-5$ with respect to $x$ is $y'=\frac{d}{dx}(3x^{2}-5)=6x$.

Step2: Use the arc - length formula

The arc - length formula for a function $y = f(x)$ on the interval $[a,b]$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. Here, $a = 0$, $b = 1$ and $y'=6x$, so $L=\int_{0}^{1}\sqrt{1+(6x)^{2}}dx=\int_{0}^{1}\sqrt{1 + 36x^{2}}dx$.

Step3: Use the integral formula for $\sqrt{a^{2}+u^{2}}$

The integral formula for $\int\sqrt{a^{2}+u^{2}}du=\frac{u}{2}\sqrt{a^{2}+u^{2}}+\frac{a^{2}}{2}\ln(u+\sqrt{a^{2}+u^{2}})+C$. In our case, $a = 1$ and $u = 6x$, so $\int\sqrt{1 + 36x^{2}}dx=\frac{6x}{2}\sqrt{1 + 36x^{2}}+\frac{1}{2}\ln(6x+\sqrt{1 + 36x^{2}})+C = 3x\sqrt{1 + 36x^{2}}+\frac{1}{2}\ln(6x+\sqrt{1 + 36x^{2}})+C$.

Step4: Evaluate the definite integral

$L=\left[3x\sqrt{1 + 36x^{2}}+\frac{1}{2}\ln(6x+\sqrt{1 + 36x^{2}})\right]_{0}^{1}$. $L=(3\times1\times\sqrt{1 + 36}+\frac{1}{2}\ln(6+\sqrt{1 + 36}))-(3\times0\times\sqrt{1+0}+\frac{1}{2}\ln(0+\sqrt{1+0}))$. $L = 3\sqrt{37}+\frac{1}{2}\ln(6+\sqrt{37})-\frac{1}{2}\ln(1)$. Since $\ln(1)=0$, $L = 3\sqrt{37}+\frac{1}{2}\ln(6+\sqrt{37})$.

Answer:

$3\sqrt{37}+\frac{1}{2}\ln(6+\sqrt{37})$