Definite Integration
Grade 12
Question:
<p>Let <span class="math-tex">\(f: {R} \rightarrow {R}\)</span> be a twice differentiable function such that <span class="math-tex">\(f(2)=1\)</span>. If <span class="math-tex">\({F}(x)=x f(x)\)</span> for all, <span class="math-tex">\(x \in {R}, \int_{0}^{2} x F^{\prime}(x) d x=6\)</span> and <span class="math-tex">\(\int_{0}^{2} x^{2} F^{\prime \prime}(x) d x=40\)</span>, then <span class="math-tex">\(F^{\prime}(2)+\int_{0}^{2} F(x) d x\)</span> is equal to:</p>
<p style="display:inline">11</p>
<p style="display:inline">15</p>
<p style="display:inline">13</p>
<p style="display:inline">9</p>
Step-by-Step Solution
Key Concept: The solution employs Integration by Parts (IBP) to transform integrals of derivatives into expressions involving the function's boundary values.
<p><span class="math-tex">$\int_{0}^{2} x F^{\prime}(x) d x=6$</span><br />
<span class="math-tex">$\left.\Rightarrow x F(x)\right|_{0} ^{2}-\int_{0}^{2} F(x) d x=6$</span><br />
<span class="math-tex">$\Rightarrow 2 F(2)-\int_{0}^{2} F(x) d x=6$</span><br />
<span class="math-tex">$[\therefore F(2)=2 f(2)=2]$</span><br />
<span class="math-tex">$\Rightarrow \int_{0}^{2} F(x) d x=-2$</span> ...(i)<br />
Also<br />
<span class="math-tex">$\int_{0}^{2} x^{2} F^{\prime \prime}(x) d x=\left.x^{2} F^{\prime}(x)\right|_{0} ^{2} $</span><span class="math-tex">$-2 \int_{0}^{2} x F^{\prime}(x) d x$</span> = 40<br />
<span class="math-tex">$\Rightarrow 4 F^{\prime}(2)-2 \times 6=40$</span><br />
<span class="math-tex">$\Rightarrow 4 F^{\prime \prime}(2)=52 \Rightarrow {~F}^{\prime}(2)=13$</span><br />
<span class="math-tex">$\therefore F^{\prime}(2)+\int_{0}^{2} F(x) d x=13-2=11$</span></p>
Correct Answer: A