Definite Integration
Grade 12
Question:
<p>Let <span class="math-tex">\(f(x)=\left\{\begin{array}{cc}-2, & -2 \leq x \leq 0 \\ x-2, & 0 \lt x \leq 2\end{array}\right.\)</span> and <span class="math-tex">\(h(x)=f(|x|)+|f(x)|\)</span>.<br />
Then <span class="math-tex">\(\int_{-2}^{2} h(x) d x\)</span> is equal to:</p>
<p style="display:inline">6</p>
<p style="display:inline">1</p>
<p style="display:inline">2</p>
<p style="display:inline">4</p>
Step-by-Step Solution
Key Concept: The integration is solved by redefining h(x) as a piecewise function through a careful domain-based evaluation of the composite function f(|x|) and the absolute value |f(x)|.
<p><img src="https://media-mycbseguide.s3.amazonaws.com/images/question_images/1775125362-kt97ap.jpg" style="height:258px; width:350px" /></p>
<p><span class="math-tex">$\therefore h(x)=\left\{\begin{array}{lc}-x-2+2=-x & -2 \leq x<0 \\ x-2+2-x=0, & 0 \leq x \leq 2\end{array}\right.$</span></p>
<p><img src="https://media-mycbseguide.s3.amazonaws.com/images/question_images/1775125412-m3nfrz.jpg" style="height:165px; width:250px" /></p>
<p><span class="math-tex">$ \therefore \int\limits_{-2}^2 h(x) d x=\int\limits_{-2}^0 h(x) d x+\int\limits_0^2 h(x) d x $</span><br />
<span class="math-tex">$ =0+2=2 $</span></p>
Correct Answer: C