Question

Find the area under the graph of \( f \) over the interval \( [-2,3] \). \[ f(x)=\left\{\begin{array}{ll} x^{2}+2 & x \leq 1 \\ 3 x & x>1 \end{array}\right. \] The area is . (Simplify your answer.)

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Answer 1

21

Explanation:

Given function:

`f(x)=\begin{cases} x^(2)+2 & x \leq 1 \\ 3 x & x > 1 \end{cases}`

To find the area under the graph of f over the interval [-2, 3], we need to split the interval [-2, 3] into two parts since the function f(x) is defined differently in these two subintervals.

To find the area under a curve over the interval [a, b], calculate the definite integral.

`\boxed{\begin{minipage}{8.5 cm}\underline{De\:\!finite integration}\\\\$\displaystyle \int^b_a \text{f}(x)\; \text{d}x=\left[\text{g}(x)\right]^b_a=\text{g}(b)-\text{g}(a)$\\\\\\where $a$ is the lower limit and $b$ is the upper limit.\\\end{minipage}}`

Therefore, the area under the graph of f over the interval [-2, 3] is given by:

`\displaystyle A=\int^1_(-2)(x^2+2)\; \text{d}x+\int^3_1 3x\; \text{d}x`

Evaluate the integral, using the following rules for integration:

`\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=(x^(n+1))/(n+1)+\text{C}$\end{minipage}}`

`\boxed{\begin{minipage}{5.1 cm}\underline{Integrating a constant}\\\\$\displaystyle \int n\:\text{d}x=nx+\text{C}$\\\\(where $n$ is any constant value) \end{minipage}}`

Therefore:

`A=\left[(x^3)/(3)+2x\right]^1_(-2)+\left[(3x^2)/(2)\right]^3_1`

`A=\left[\left(((1)^3)/(3)+2(1)\right)-\left(((-2)^3)/(3)+2(-2)\right)\right]+\left[\left((3(3)^2)/(2)\right)-\left((3(1)^2)/(2)\right)\right]`

`A=\left[\left((1)/(3)+2\right)-\left((-8)/(3)-4\right)\right]+\left[\left((27)/(2)\right)-\left((3)/(2)\right)\right]`

`A=(7)/(3)+(20)/(3)+(27)/(2)-(3)/(2)`

`A=(27)/(3)+(24)/(2)`

`A=9+12`

`A=21`

Therefore, the total area under the graph of f(x) over the interval [-2, 3] is 21 square units.