Consider the following statements (A) and (B) (A) $\int_a^b \frac{d}{d x}(f(x)) d x=\frac{d}{d x}…
Consider the following statements (A) and (B)
(A) $\int_a^b \frac{d}{d x}(f(x)) d x=\frac{d}{d x} \int_a^b(f(x)) d x$
(B) $\frac{d}{d x}\left(\int f(x) d x\right)=f(x)+C$
Which one of the following is true?
Only (A) is true
Only (B) is true
Both $(\mathrm{A})$ and $(\mathrm{B})$ are true
Both (A) and (B) are false
Solution
$\int_a^b \frac{d}{d x}(f(x)) d x=\int_a^b d(f(x))=f(b)-f(a)$ ...(i)
and $\frac{d}{d x} \int_a^b f(x) d x=\frac{d}{d x}[\mathrm{~A}$ constant $]=0$ ...(ii)
$\left\{\because \int_a^b f(x) d x\right.$ will necessarily give a constant value
Hence $\int_a^b \frac{d}{d x}(f(x)) d x \neq \frac{d}{d x} \int_a^b f(x) d x$
Therefore statement ' $\mathrm{A}$ ' is false.
Now consider
$
\frac{d}{d x}\left(\int f(x) d x\right)=\frac{d}{d x}\left(g(x)+C_1\right)
$
where $\int f(x) d x=g(x)+C_1$
$
\begin{aligned}
& \Rightarrow \frac{d}{d x} \int f(x) d x=g^{\prime}(\mathrm{x})+0\left\{\because \frac{d}{d x}(\text { constant })=0\right. \\
& =g^{\prime}(x)
\end{aligned}
$
Clearly statement ' $\mathrm{B}$ ' is also false