Definite Integration Question 19
Question 19
- Let $a+b=4$, where $a<2$ and let $g(x)$ be a differentiable function. If $\frac{d g}{d x}>0, \forall x$ prove that $\int_{0}^{a} g(x) d x+\int_{0}^{b} g(x) d x$ increases as $(b-a) \underset{(1997,5 \mathrm{M})}{\text { increases. }}$
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Answer:
Correct Answer: 19. $(n=3)$
Solution:
- Let $t=b-a$ and $a+b=4$
[given]
$$ \begin{aligned} & \Rightarrow \quad t=4-a-a \ & \Rightarrow \quad t=4-2 a \ & \Rightarrow \quad a=2-\frac{t}{2} \ & \text { and } \quad t=b-(4-b) \ & \Rightarrow \quad t=2 b-4 \ & \Rightarrow \quad \frac{t}{2}=b-2 \ & \Rightarrow \quad b=2+\frac{t}{2} \end{aligned} $$
Again, $\quad a<2$
$$ \begin{array}{ll} \Rightarrow & 2-\frac{\pi}{2}<2 \ \Rightarrow & \frac{\pi}{2}>0 \Rightarrow t>0 \end{array} $$
[given]
Now, $\int_{0}^{a} g(x) d x+\int_{0}^{b} g(x) d x$
$$ =\int_{0}^{2-t / 2} g(x) d x+\int_{0}^{2+t / 2} g(x) d x $$
Let $\quad F(x)=\int_{0}^{2-t / 2} g(x) d x+\int_{0}^{2+t / 2} g(x) d x$
For $t>0, F^{\prime}(t)=-\frac{1}{2} g \quad 2-\frac{t}{2}+\frac{1}{2} g \quad 2+\frac{t}{2}$
[using Leibnitz’s rule]
$$ =\frac{1}{2} g \quad 2+\frac{t}{2}-\frac{1}{2} g \quad 2-\frac{t}{2} $$
Again, $\quad \frac{d g}{d x}>0, \forall x \in R$
[given]
Now, $2-t / 2<2+t / 2 \therefore t>0$
We get $g(2+t / 2)-g(2-t / 2)>0, \forall t>0$
So, $\quad F^{\prime}(t)>0, \forall t>0$
Hence, $F(t)$ increases with $t$, therefore $F(t)$ increases as $(b-a)$ increases.
