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If $f(x)=\int_x^{x+1} e^{-t^2} d t$, then the interval in which $f(x)$ is decreasing is
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Verified Answer
The correct answer is:
$\left(-\frac{1}{2}, \infty\right)$
We have,
$$
\begin{aligned}
& f(x)=\int_x^{x+1} e^{-t^2} d t \Rightarrow f^{\prime}(x)=e^{-(x+1)^2}-e^{-x^2} \\
& f^{\prime}(x)=\frac{1}{e^{(x+1)^2}}-\frac{1}{e^{x^2}}
\end{aligned}
$$
Since, $f(x)$ is decreasing function.
$$
\begin{array}{ll}
\therefore & f^{\prime}(x) < 0 \\
\therefore \quad & \frac{1}{e^{(x+1)^2}}-\frac{1}{e^{x^2}} < 0 \Rightarrow e^{x^2} < e^{(x+1)^2} \\
& x^2 < (x+1)^2 \Rightarrow x^2+2 x+1-x^2>0 \\
& 2 x+1>0 \Rightarrow x>-\frac{1}{2} \\
\therefore & x \in\left(\frac{-1}{2}, \infty\right)
\end{array}
$$
$$
\begin{aligned}
& f(x)=\int_x^{x+1} e^{-t^2} d t \Rightarrow f^{\prime}(x)=e^{-(x+1)^2}-e^{-x^2} \\
& f^{\prime}(x)=\frac{1}{e^{(x+1)^2}}-\frac{1}{e^{x^2}}
\end{aligned}
$$
Since, $f(x)$ is decreasing function.
$$
\begin{array}{ll}
\therefore & f^{\prime}(x) < 0 \\
\therefore \quad & \frac{1}{e^{(x+1)^2}}-\frac{1}{e^{x^2}} < 0 \Rightarrow e^{x^2} < e^{(x+1)^2} \\
& x^2 < (x+1)^2 \Rightarrow x^2+2 x+1-x^2>0 \\
& 2 x+1>0 \Rightarrow x>-\frac{1}{2} \\
\therefore & x \in\left(\frac{-1}{2}, \infty\right)
\end{array}
$$
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