What is the area of the region between the graphs of f(x)=sqrt(x+1) and g(x)=2x-4 from x=0 to x=3 ? Choose 1 answer: (A) (5)/(3) (B) -3 (C) (14)/(3) (D) (23)/(3)

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What is the area of the region between the graphs of  f(x)=sqrt(x+1) and  g(x)=2x-4 from  x=0 to  x=3 ? Choose 1 answer: (A)  (5)/(3) (B) -3 (C)  (14)/(3) (D)  (23)/(3)

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Set up integral for area: To find the area between the two curves, we need to integrate the difference between the functions from x=0 to x=3. First, we need to set up the integral for the area between the curves. The area A is given by the integral from 0 to 3 of the top function minus the bottom function. Since we are looking for the area between the curves, we need to determine which function is on top (higher value) for the interval from x=0 to x=3.Determine top function: We evaluate both functions at a point within the interval to determine which function is on top. Let's choose x=1 for simplicity. f(1) = sqrt(1+1) = sqrt(2) g(1) = 2(1)-4 = -2 Since sqrt(2) > -2, f(x) is the top function and g(x) is the bottom function on the interval from x=0 to x=3.Set up integral equation: Now we can set up the integral to find the area between the curves. A = ∫ from 0 to 3 of (f(x) - g(x)) dx A = ∫ from 0 to 3 of (sqrt(x+1) - (2x-4)) dxCalculate the integral: Next, we calculate the integral. A = ∫ from 0 to 3 of (sqrt(x+1) - 2x + 4) dx This requires us to integrate term by term.Integrate sqrt(x+1): We integrate the first term, sqrt(x+1), using a substitution if necessary. Let u = x+1, then du = dx, and when x=0, u=1, and when x=3, u=4. The integral of sqrt(u) with respect to u is (2/3)u^(3/2).Integrate -2x: We integrate the second term, -2x, which is a simple power function. The integral of -2x with respect to x is -x^2.Integrate +4: We integrate the third term, +4, which is a constant. The integral of 4 with respect to x is 4x.Combine and evaluate: Now we combine the integrated terms and evaluate from 0 to 3. A = [(2/3)(x+1)^(3/2) - x^2 + 4x] from 0 to 3Evaluate upper limit: We evaluate the expression at the upper limit, x=3. A(3) = (2/3)(3+1)^(3/2) - 3^2 + 4*3 A(3) = (2/3)(4)^(3/2) - 9 + 12 A(3) = (2/3)(8) - 9 + 12 A(3) = (16/3) - 9 + 12 A(3) = (16/3) + 3Evaluate lower limit: We evaluate the expression at the lower limit, x=0. A(0) = (2/3)(0+1)^(3/2) - 0^2 + 4*0 A(0) = (2/3)(1) - 0 + 0 A(0) = 2/3Find the area: We subtract the value of the expression at the lower limit from the value at the upper limit to find the area. A = A(3) - A(0) A = [(16/3) + 3] - (2/3) A = (16/3) + (9/3) - (2/3) A = (16/3) + (7/3) A = (23/3)

What is the area of the region between the graphs of $f(x)=\sqrt{x+1}$ and $g(x)=2 x-4$ from $x=0$ to $x=3$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{5}{3}$ $\newline$ (B) $-3$ $\newline$ (C) $\frac{14}{3}$ $\newline$ (D) $\frac{23}{3}$

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