Question:

Step 1: Recognize the structure of the integrand. We need to evaluate $I = \displaystyle\int_{-1}^{0} x f'(x) f(x)\,dx$. Notice that the product $f'(x)f(x)$ appears in the integrand. We can rewrite this using the chain rule: $$f'(x)f(x) = \frac{1}{2}\frac{d}{dx}[f(x)]^2$$ This is because $\frac{d}{dx}[f(x)]^2 = 2f(x)f'(x)$. Step 2: Rewrite the integral using the derivative form. Substituting the expression from Step 1: $$I = \int_{-1}^{0} x \cdot \frac{1}{2}\frac{d}{dx}[f(x)]^2\,dx$$ Step 3: Apply integration by parts. We use integration by parts with: - $u = x$, so $du = dx$ - $dv = \frac{1}{2}d[f(x)]^2$, so $v = \frac{1}{2}[f(x)]^2$ Using the integration by parts formula $\int u\,dv = uv - \int v\,du$: $$I = \left[\frac{x}{2}(f(x))^2\right]_{-1}^{0} - \int_{-1}^{0} \frac{1}{2}(f(x))^2\,dx$$ Step 4: Evaluate the boundary term. The boundary term is: $$\left[\frac{x}{2}(f(x))^2\right]_{-1}^{0} = \frac{0}{2}(f(0))^2 - \frac{(-1)}{2}(f(-1))^2$$ $$= 0 + \frac{1}{2}(f(-1))^2$$ Since $f(-1) = 2$: $$= \frac{1}{2}(2)^2 = \frac{1}{2} \cdot 4 = 2$$ Step 5: Evaluate the remaining integral term. The second term is: $$\int_{-1}^{0} \frac{1}{2}(f(x))^2\,dx = \frac{1}{2}\int_{-1}^{0}(f(x))^2\,dx$$ Given that $\displaystyle\int_{-1}^{0}(f(x))^2\,dx = 10$: $$= \frac{1}{2} \cdot 10 = 5$$ Step 6: Combine the results to find the final answer. $$I = 2 - 5 = -3$$ Therefore, $\displaystyle\int_{-1}^{0} x f'(x) f(x)\,dx = -3$. The answer is **Option 3: $-3$** <div class="key-concept"><strong>Key Concept:</strong> Integration by parts combined with the identity $f'(x)f(x) = \frac{1}{2}\frac{d}{dx}[f(x)]^2$</div> <div class="trap-box"><strong>Trap:</strong> Not recognizing that $f'(x)f(x) = \frac{1}{2}\frac{d}{dx}[f(x)]^2$ which simplifies the integration by parts significantly.</div>