Given the function \( f(x)=12 x^{5}-4 x^{3}+2 x \) for \( x \) between -1 and 2, find: The net area \( =[?] \) The total area \( =[ \)

We are given \[ f(x)=12x^5-4x^3+2x \] for \( x \) in the interval \([-1,2]\). ### Step 1. Find the antiderivative We integrate term‐by‐term: \[ \int 12x^5\,dx = 2x^6, \] \[ \int (-4x^3)\,dx = -x^4, \] \[ \int 2x\,dx = x^2. \] Thus, an antiderivative \( F(x) \) is \[ F(x)=2x^6 - x^4 + x^2. \] ### Step 2. Compute the Net Area The net area is given by \[ \int_{-1}^{2} f(x)\,dx = F(2)-F(-1). \] First, compute \( F(2) \): \[ F(2)=2(2)^6 - (2)^4 + (2)^2 = 2\cdot64 - 16 + 4 = 128 - 16 + 4 = 116. \] Next, compute \( F(-1) \): \[ F(-1)=2(-1)^6 - (-1)^4 + (-1)^2 =2\cdot1 - 1+1 = 2-1+1=2. \] Thus, the net area is \[ F(2)-F(-1)=116-2=114. \] ### Step 3. Determine Where the Function Changes Sign We need the total area, which is the sum of the absolute areas where \( f(x) \) may be negative. Notice that \[ f(x)=12x^5-4x^3+2x = x\left(12x^4-4x^2+2\right). \] The factor \(12x^4-4x^2+2\) is always positive for all real \( x \) (its discriminant is negative when viewed as a quadratic in \( x^2 \)). Hence, the sign of \( f(x) \) is determined by the factor \( x \): - For \( x<0 \), \( f(x)<0 \). - For \( x>0 \), \( f(x)>0 \). Thus, the only sign change in the interval \([-1,2]\) is at \( x=0 \). ### Step 4. Compute the Total Area The total area is \[ \text{Total Area}=\int_{-1}^{0} \left|f(x)\right|\,dx+\int_{0}^{2} \left|f(x)\right|\,dx. \] Since \( f(x) \) is negative on \([-1,0]\) and positive on \([0,2]\): \[ \int_{-1}^{0} \left|f(x)\right|\,dx = -\int_{-1}^{0} f(x)\,dx, \] \[ \int_{0}^{2} \left|f(x)\right|\,dx = \int_{0}^{2} f(x)\,dx. \] #### Compute \(\int_{-1}^{0} f(x)\,dx\) \[ \int_{-1}^{0} f(x)\,dx=F(0)-F(-1)=0-2=-2. \] Taking the absolute value gives \( 2 \). #### Compute \(\int_{0}^{2} f(x)\,dx\) \[ \int_{0}^{2} f(x)\,dx=F(2)-F(0)=116-0=116. \] Therefore, the total area is \[ 2+116=118. \] ### Final Answers Net area: \[ 114 \] Total area: \[ 118 \]