Review question

# When is $I(c) = \int^1_0 2^{-(x-c)^2} dx$ largest? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7461

## Solution

1. The graph $y = f(x)$ of a certain function has been plotted below.

On the next three pairs of axes (A), (B), (C) are graphs of $y = f(-x), \quad f(x-1), \quad -f(x)$ in some order. Say which axes correspond to which graphs.

Graph (A) is $-f(x)$, since the graph has been reflected in the $x$-axis.

Graph (B) is $f(-x)$, since the graph has been reflected in the $y$-axis.

Graph (C) is $f(x-1)$, since the graph has been translated (one unit) to the right.

1. Sketch graphs of both of the following functions $y = 2^{-x^2} \quad \text{and} \quad y = 2^{2x - x^2}.$ Carefully label any stationary points.

We sketch $y = 2^{-x^2}$ by noticing that it must be symmetrical about the $y$-axis. It takes its largest value of $1$ at $x = 0$, and decreases to zero as the magnitude of $x$ increases.

For the second graph, we note that $y = 2^{2x - x^2} = 2^{1-1 + 2x - x^2} = 2^{1-(x-1)^2}= 2 \times 2^{-(x-1)^2},$ and so this graph is $y = 2^{-x^2}$ translated to the right by one unit and stretched vertically by a factor of two.

1. Let $c$ be a real number and define the following integral $I(c) = \int^1_0 2^{-(x-c)^2} dx.$

State the value(s) of $c$ for which $I(c)$ is largest. Briefly explain your reasoning. [Note you are not being asked to calculate this maximum value.]

The graph of $2^{-(x-c)^2}$ is the graph of $2^{-x^2}$ translated $c$ units to the right.

The integral $I(c)$ corresponds to the area under that graph within the range $0 \le x \le 1$.

To maximise this, we should choose $c$ so that the highest point of the graph is in the middle of this range, at $x = \dfrac{1}{2}$, which occurs when $c = \dfrac{1}{2}$.