Question Statement

Evaluate the integrals in each part when $f(x)=\{\begin{array}{ll}x& \text{if }x\le 1\\ 3& \text{if }x>1\end{array}$

(i) ${\int }_0^{1}f(x)dx$

(ii) ${\int }_{-1}^{1}f(x)dx$

(iii) ${\int }_1^{4}f(x)dx$

(iv) ${\int }_{-1}^{2}f(x)dx$

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Background and Explanation

This problem requires integrating a piecewise function, where the definition of $f(x)$ changes at $x=1$. The key is to identify which expression for $f(x)$ applies to the interval of integration. When an interval crosses the boundary point, we use the additive property of definite integrals to split the integral into separate parts.

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Solution

Part (i): ${\int }_0^{1}f(x)dx$

For the interval $[0,1]$, we have $x\le 1$, so we use $f(x)=x$.

$\begin{array}{rl}{\int }_0^{1}f(x)dx& ={\int }_0^{1}xdx\\ & ={\left|\frac{x^2}{2}\right|}_0^1\\ & =\frac{1}{2}\left((1{)}^2-(0{)}^2\right)\\ & =\frac{1}{2}(1-0)\\ & =\frac{1}{2}\end{array}$

Part (ii): ${\int }_{-1}^{1}f(x)dx$

For the interval $[-1,1]$, all values satisfy $x\le 1$, so we use $f(x)=x$.

$\begin{array}{rl}{\int }_{-1}^{1}f(x)dx& ={\int }_{-1}^{1}xdx\\ & ={\left|\frac{x^2}{2}\right|}_{-1}^1\\ & =\frac{1}{2}\left((1{)}^2-(-1{)}^2\right)\\ & =\frac{1}{2}(1-1)\\ & =\frac{1}{2}(0)\\ & =0\end{array}$

Part (iii): ${\int }_1^{4}f(x)dx$

For the interval $[1,4]$, we use the definition $f(x)=3$ (valid for $x>1$). Note that the single point $x=1$ where $f(1)=1$ does not affect the value of the definite integral.

$\begin{array}{rl}{\int }_1^{4}f(x)dx& ={\int }_1^{4}3dx\\ & =3{\int }_1^{4}1dx\\ & =3{\left|x\right|}_1^4\\ & =3(4-1)\\ & =3(3)\\ & =9\end{array}$

Part (iv): ${\int }_{-1}^{2}f(x)dx$

The interval $[-1,2]$ crosses the boundary at $x=1$. We split the integral into two parts: from $-1$ to $1$ (where $f(x)=x$) and from $1$ to $2$ (where $f(x)=3$).

$\begin{array}{rl}{\int }_{-1}^{2}f(x)dx& ={\int }_{-1}^{1}f(x)dx+{\int }_1^{2}f(x)dx\\ & ={\int }_{-1}^{1}xdx+{\int }_1^{2}3dx\\ & ={\left|\frac{x^2}{2}\right|}_{-1}^1+3{\left|x\right|}_1^2\\ & =\frac{1}{2}\left((1{)}^2-(-1{)}^2\right)+3(2-1)\\ & =\frac{1}{2}(1-1)+3(1)\\ & =\frac{1}{2}(0)+3\\ & =0+3\\ & =3\end{array}$

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Key Formulas or Methods Used

• Power Rule for Integration: $\int xdx=\frac{x^2}{2}+C$
• Constant Rule for Integration: $\int kdx=kx+C$ (where $k$ is a constant)
• Additivity Property: ${\int }_a^{c}f(x)dx={\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx$ for splitting integrals at boundary points
• Fundamental Theorem of Calculus: ${\left|F(x)\right|}_a^b=F(b)-F(a)$ for evaluating definite integrals

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Summary of Steps

1. Identify the breakpoint: Locate where the function definition changes ($x=1$).
2. Analyze the interval: Check if the integration interval lies entirely in $x\le 1$, entirely in $x>1$, or crosses both.
3. Select the appropriate function: Use $f(x)=x$ for intervals in $x\le 1$, and $f(x)=3$ for intervals in $x>1$.
4. Split if necessary: For intervals crossing $x=1$, break the integral into two separate integrals at $x=1$.
5. Integrate and evaluate: Apply the power rule or constant rule, then use the Fundamental Theorem of Calculus to compute the numerical value.