Use the limit process to find the area of the region. y=(1)/(4)x^(3),[2,4]

Set up integral: First, we need to set up the integral for the area under the curve from x=2 to x=4. We use the formula for the definite integral of y = (1/4)x^3 from x = 2 to x = 4. Integral setup: ∫ from 2 to 4 of (1/4)x^3 dx Calculate antiderivative: Next, we calculate the antiderivative of (1/4)x^3. Antiderivative of (1/4)x^3 is (1/4)*(1/4)x^4 = (1/16)x^4. Now, we evaluate this from x=2 to x=4. Evaluate integral: Plug in the upper and lower limits of the integral. Evaluate at x=4: (1/16)(4^4) = (1/16)(256) = 16. Evaluate at x=2: (1/16)(2^4) = (1/16)(16) = 1. Subtract the lower limit evaluation from the upper limit evaluation: 16 - 1 = 15.

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Use the limit process to find the area of the region. $\newline$ $y=\frac{1}{4}x^{3},[2,4]$

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Q. Use the limit process to find the area of the region.

\newline

y=\frac{1}{4}x^{3},[2,4]

Set up integral: First, we need to set up the integral for the area under the curve from $x=2$ to $x=4$ . $\newline$ We use the formula for the definite integral of $y = \frac{1}{4}x^3$ from $x = 2$ to $x = 4$ . $\newline$ Integral setup: $\int_{2}^{4} \frac{1}{4}x^3 \, dx$
Calculate antiderivative: Next, we calculate the antiderivative of $(\frac{1}{4})x^3$ . Antiderivative of $(\frac{1}{4})x^3$ is $(\frac{1}{4})\cdot(\frac{1}{4})x^4 = (\frac{1}{16})x^4$ . Now, we evaluate this from $x=2$ to $x=4$ .
Evaluate integral: Plug in the upper and lower limits of the integral. $\newline$ Evaluate at $x=4$ : $(1/16)(4^4) = (1/16)(256) = 16$ . $\newline$ Evaluate at $x=2$ : $(1/16)(2^4) = (1/16)(16) = 1$ . $\newline$ Subtract the lower limit evaluation from the upper limit evaluation: $16 - 1 = 15$ .