Definite Integration Question 54
Question 54
- Let $f:[1, \infty] \rightarrow[2, \infty]$ be differentiable function such that $f(1)=2$. If $6 \int_{1}^{x} f(t)=d t=3 x f(x)-x^{3}, \forall x \geq 1$ then the value of $f(2)$ is
(2011)
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Solution:
- Given, $f(1)=\frac{1}{3}$ and $6 \int_{1}^{x} f(t) d t=3 x f(x)-x^{3}, \forall x \geq 1$
Using Newton-Leibnitz formula.
Differentiating both sides
$$ \begin{aligned} & \Rightarrow \quad 6 f(x) \cdot 1-0=3 f(x)+3 x f^{\prime}(x)-3 x^{2} \ & \Rightarrow \quad 3 x f^{\prime}(x)-3 f(x)=3 x^{2} \Rightarrow \quad f^{\prime}(x)-\frac{1}{x} f(x)=x \ & \Rightarrow \quad \frac{x f^{\prime}(x)-f^{\prime}(x)}{x^{2}}=1 \Rightarrow \quad \frac{d}{d x} \frac{x}{x}=1 \end{aligned} $$
On integrating both sides, we get
$$ \begin{aligned} \Rightarrow \quad \frac{f(x)}{x} & =x+c \ \frac{1}{3} & =1+c \Rightarrow c=\frac{2}{3} \text { and } f(x)=x^{2}-\frac{2}{3} x \ \therefore \quad f(2) & =4-\frac{4}{3}=\frac{8}{3} \end{aligned} $$
NOTE Here, $f(1)=2$, does not satisfy given function.
$$ \therefore \quad f(1)=\frac{1}{3} $$
For that $f(x)=x^{2}-\frac{2}{3} x$ and $f(2)=4-\frac{4}{3}=\frac{8}{3}$
