Definite Integration
Grade None
Question:
<p>The minimum value of the twice differentiable function f(x) = <span class="math-tex">\(\int \limits_0^x e^{x-t} f^{\prime}(t) d t-\left(x^2-x+1\right) e^x\)</span>, <span class="math-tex">\( x \in R\)</span>, is:</p>
<p style="display:inline"><span class="math-tex">\(-2 \sqrt{e}\)</span></p>
<p style="display:inline"><span class="math-tex">\(-\sqrt{e}\)</span></p>
<p style="display:inline"><span class="math-tex">\(-\frac{2}{\sqrt{e}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{\sqrt{e}}\)</span></p>
Step-by-Step Solution
Key Concept: Factor out the $e^x$ term from the integral to simplify the expression before differentiating via Leibniz's Rule to determine the functional form of $f(x)$.
<p><span class="math-tex">$f(x)=e^x \cdot \int \limits_0^x \frac{f^{\prime}(t)}{e^t} d t$</span><br />
<span class="math-tex">$f^{\prime}(x)=e^x \cdot \int\limits_0^x \frac{f^{\prime}(t)}{e^t} d t+e^x$</span><span class="math-tex">$-\left[(2 x-1) \cdot e^x+\left(x^2-x+1\right) \cdot e^x\right]$</span><br />
Now <span class="math-tex">$\int\limits_0^x \frac{f^{\prime}(t)}{e^t} d t=x^2+x$</span><br />
Differentiate on both sides w.r.t 'x'<br />
<span class="math-tex">$\Rightarrow \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{e}^{\mathrm{x}}}=2 \mathrm{x}+1 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})$</span> <span class="math-tex">$=(2 \mathrm{x}+1) \cdot \mathrm{e}^{\mathrm{x}}$</span><br />
Now <span class="math-tex">$ f^{\prime}(x)=0 \Rightarrow x=-\frac{1}{2}$</span><br />
<span class="math-tex">$f(x)=(2 x+1) \cdot e^x-2 e^x+C$</span><br />
<span class="math-tex">$\therefore$</span> f (0) = -1<br />
-1 = 1 - 2 + C <span class="math-tex">$\Rightarrow$</span> c = 0<br />
Now <span class="math-tex">$f(x)=e^x(2 x-1)$</span><br />
<span class="math-tex">$\Rightarrow f\left(-\frac{1}{2}\right)=\frac{-2}{\sqrt{e}}$</span></p>
Correct Answer: C