How Translations Affect Enclosed Area

Introduction

A translation moves a graph without changing its shape. If both boundary graphs move together by the same amount, the region between them moves too, but its area does not change. If only one boundary moves, however, the gap between the curves and their intersection points can change.

The key is to distinguish a shared translation of the entire region from a translation of just one curve. In either case, area must be calculated using the boundaries and endpoints that apply after the translation.

Writing a Translated Graph

Suppose a graph is given by $y=f(x)$. Translating it horizontally by $h$ and vertically by $k$ gives the graph

$$ y=f(x-h)+k. $$

The subtraction inside the function is important: $x-h$ shifts the graph to the right by $h$ when $h>0$. The added $k$ shifts it up by $k$ when $k>0$. Negative values shift in the opposite directions.

Definition: A translation by $(h,k)$ sends each point $(x,y)$ to $(x+h,y+k)$. Thus the graph $y=f(x)$ becomes $y=f(x-h)+k$.

For two boundary graphs $y=f(x)$ and $y=g(x)$, their common translation by $(h,k)$ is therefore

y=f(x-h)+k \qquad\text{and}\qquad y=g(x-h)+k.

If the original region extends from $x=a$ to $x=b$, the translated region extends from $x=a+h$ to $x=b+h$. Both the graphs and the region's horizontal location have changed.

Why a Common Translation Preserves Area

When one continuous graph is above another on an interval $[a,b]$, the area between them is the integral of the upper function minus the lower function. If the order changes, geometric area is found by integrating the absolute difference, or by splitting the interval at intersections. In either form, a common translation leaves the vertical separation unchanged.

Theorem: If a finite region between $y=f(x)$ and $y=g(x)$ is translated by the same vector $(h,k)$, its area is unchanged. In the integral description, the translated area is $$ \int_{a+h}^{b+h}\left|f(x-h)+k-\bigl(g(x-h)+k\bigr)\right|\,dx = \int_a^b |f(u)-g(u)|\,du. $$

Here $u=x-h$, so $dx=du$, and the endpoints become $u=a$ and $u=b$. The vertical translation $k$ cancels from the difference. This argument also applies if the curves cross: the absolute difference accounts for the change in which graph is higher.

Translate both boundaries.
Replace each $f(x)$ by $f(x-h)+k$ and $g(x)$ by $g(x-h)+k$.

Track the region.
Shift the horizontal endpoints from $a,b$ to $a+h,b+h$.

Compare the gaps.
The common $k$ cancels, and the substitution $u=x-h$ returns the original area integral.

Worked Example: Translating Both Boundaries

Find the area enclosed by $y=x^2$ and $y=2$, then translate both graphs right by $1$ and up by $3$. Verify that the area is unchanged.

First find the original intersections:

x^2=2 \quad\Longrightarrow\quad x=-\sqrt{2},\ \sqrt{2}.

On this interval, $2$ is above $x^2$. Thus

A=\int_{-\sqrt{2}}^{\sqrt{2}}(2-x^2)\,dx =\left[2x-\frac{x^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}}.

At $x=\sqrt{2}$, the antiderivative is $2\sqrt{2}-2\sqrt{2}/3=4\sqrt{2}/3$. At $x=-\sqrt{2}$, it is $-2\sqrt{2}+2\sqrt{2}/3=-4\sqrt{2}/3$. Therefore

A=\frac{4\sqrt{2}}{3}-\left(-\frac{4\sqrt{2}}{3}\right) =\frac{8\sqrt{2}}{3}.

After translating both graphs right by $1$ and up by $3$, their equations are $y=(x-1)^2+3$ and $y=5$. Their intersections satisfy $(x-1)^2+3=5$, so $x=1-\sqrt{2}$ and $x=1+\sqrt{2}$. The area is

\begin{aligned} A_{\mathrm{translated}} &=\int_{1-\sqrt{2}}^{1+\sqrt{2}}\left(5-\bigl((x-1)^2+3\bigr)\right)\,dx\\ &=\int_{1-\sqrt{2}}^{1+\sqrt{2}}\left(2-(x-1)^2\right)\,dx. \end{aligned}

Set $u=x-1$; the endpoints become $-\sqrt{2}$ and $\sqrt{2}$. The resulting integral is the original one, so $A_{\mathrm{translated}}=8\sqrt{2}/3$. As a check, differentiating $2u-u^3/3$ gives $2-u^2$, the integrand used for the translated area.

Translating Only One Boundary

If only one graph moves, its separation from the other graph is generally no longer the same. A vertical translation changes the function values; a horizontal translation changes where the translated graph lies over each $x$-value. In both cases, intersection points may move, so they must be found again before setting up an area integral.

The area formula itself does not change: once the new intersections are known, integrate the upper boundary minus the lower boundary between them when one stays above the other. If the order changes, split the interval at the intersection or use the absolute difference.

Worked Example: Moving One Boundary Vertically

The curves $y=x^2$ and $y=4$ enclose a region. Translate only the horizontal boundary up by $1$, then find the new enclosed area.

The moved line is $y=5$. Its intersections with the unchanged parabola satisfy

x^2=5 \quad\Longrightarrow\quad x=-\sqrt{5},\ \sqrt{5}.

The line is above the parabola between these points. Therefore

\begin{aligned} A &=\int_{-\sqrt{5}}^{\sqrt{5}}(5-x^2)\,dx\\ &=\left[5x-\frac{x^3}{3}\right]_{-\sqrt{5}}^{\sqrt{5}}. \end{aligned}

At $x=\sqrt{5}$, the antiderivative equals $5\sqrt{5}-5\sqrt{5}/3=10\sqrt{5}/3$. At $x=-\sqrt{5}$, it equals $-10\sqrt{5}/3$. Hence

A=\frac{10\sqrt{5}}{3}-\left(-\frac{10\sqrt{5}}{3}\right) =\frac{20\sqrt{5}}{3}.

As a check, the derivative of $5x-x^3/3$ is $5-x^2$, the height of each vertical slice. The new area is greater than the original area $16/3$, as expected from raising the line and expanding the enclosed region.

Worked Example: Moving One Boundary Horizontally

Keep $y=x^2$ fixed and translate the upper boundary $y=4-x^2$ right by $1$. Find the area enclosed by the new pair of curves.

The translated upper boundary is $y=4-(x-1)^2$. To find the intersections, set it equal to $x^2$:

4-(x-1)^2=x^2 \quad\Longrightarrow\quad 4-(x^2-2x+1)=x^2 \quad\Longrightarrow\quad 2x^2-2x-3=0.

Using the quadratic formula gives

x=\frac{2\pm\sqrt{(-2)^2-4(2)(-3)}}{4} =\frac{2\pm\sqrt{28}}{4} =\frac{1\pm\sqrt{7}}{2}.

Subtracting the fixed lower boundary from the translated upper boundary gives $3+2x-2x^2$. Completing the square rewrites this as $7/2-2(x-1/2)^2$. With $u=x-1/2$, the endpoints become $-\sqrt{7}/2$ and $\sqrt{7}/2$, so

\begin{aligned} A &=\int_{-\sqrt{7}/2}^{\sqrt{7}/2}\left(\frac{7}{2}-2u^2\right)\,du\\ &=\left[\frac{7u}{2}-\frac{2u^3}{3}\right]_{-\sqrt{7}/2}^{\sqrt{7}/2}. \end{aligned}

Write $r=\sqrt{7}/2$. At $u=r$, the antiderivative is $7r/2-2r^3/3$. Since $r^2=7/4$, this is $7r/2-7r/6=7r/3$. At $u=-r$, its value is $-7r/3$. Thus

A=\frac{14r}{3}=\frac{14}{3}\cdot\frac{\sqrt{7}}{2} =\frac{7\sqrt{7}}{3}.

For a check, differentiating $7u/2-2u^3/3$ gives $7/2-2u^2$, which is exactly the vertical gap. The integrand is positive between the two roots, confirming the chosen upper and lower boundaries.

A Common Mistake: Keeping the Old Endpoints

After shifting $y=4$ up to $y=5$ in the previous vertical-translation example, the old intersections with $y=x^2$ are still at $x=-2$ and $x=2$. They are not intersections of the new pair: at either endpoint $x^2=4$, not $5$. Integrating between the old endpoints gives

\int_{-2}^{2}(5-x^2)\,dx =\left[5x-\frac{x^3}{3}\right]_{-2}^{2} =\frac{44}{3},

but this does not measure the region enclosed by the new curves. The correct intersections are $\pm\sqrt{5}$, giving the correct area $20\sqrt{5}/3$. The general correction is simple: after moving only one boundary, solve the new intersection equation before integrating. By contrast, a common translation moves the intersections along with the region, preserving the area.

Why Translation Matters

Translation is useful for recognizing when an area calculation can be reused. If every boundary moves together, the geometry is the same up to location, so the area is identical. If only one boundary moves, a familiar-looking graph may enclose a different region: its new equation and its new intersections determine the answer.

Takeaway: A common translation preserves the gap between the boundaries and the enclosed area. Translating just one boundary generally changes both the gap and the intersection points; recalculate the intersections before finding area.

Check Your Understanding

Try answering these questions using the translation rule and the area setup shown above.

What equation represents $y=f(x)$ translated left by $3$ and down by $2$?
Why does adding the same vertical shift to both boundary functions leave their difference unchanged?
Two curves enclose a finite region. What happens to its area when both curves are translated right by $h$ and up by $k$?
If $y=x^2$ and $y=9$ are enclosed by their intersection points, what are the endpoints after translating only $y=9$ up by $1$?
Why is it generally incorrect to use the original intersections after translating only one boundary?

How Translations Affect Enclosed Area

What You'll Learn

Introduction

Writing a Translated Graph

Why a Common Translation Preserves Area

Worked Example: Translating Both Boundaries

Worked Example: Translating Both Boundaries

Translating Only One Boundary

Worked Example: Moving One Boundary Vertically

Worked Example: Moving One Boundary Vertically

Worked Example: Moving One Boundary Horizontally

Worked Example: Moving One Boundary Horizontally

A Common Mistake: Keeping the Old Endpoints

Why Translation Matters

Check Your Understanding

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How Translations Affect Enclosed Area

What You'll Learn

Introduction

Writing a Translated Graph

Why a Common Translation Preserves Area

Worked Example: Translating Both Boundaries

Worked Example: Translating Both Boundaries

Translating Only One Boundary

Worked Example: Moving One Boundary Vertically

Worked Example: Moving One Boundary Vertically

Worked Example: Moving One Boundary Horizontally

Worked Example: Moving One Boundary Horizontally

A Common Mistake: Keeping the Old Endpoints

Why Translation Matters

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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