Area in rectangle-shaped Coordinates

Recall the the area under the graph the a consistent function \(f\left( x \right)\) in between the vertical lines \(x = a,\) \(x = b\) can be computed by the identify integral:

where \(F\left( x \right)\) is any kind of antiderivative of \(f\left( x \right).\)

Figure 1.

You are watching: Find the area of the region bounded by the graphs of the equations.

We can expand the notion of the area under a curve and consider the area of the region between 2 curves.

If \(f\left( x \right)\) and also \(g\left( x \right)\) space two constant functions and \(f\left( x \right) \ge g\left( x \right)\) top top the closed interval \(\left< a,b \right>,\) climate the area between the curves \(y = f\left( x \right)\) and \(y = g\left( x \right)\) in this expression is given by

Figure 2.

In terms of antiderivatives, the area of region is to express in the form

\dx = F\left( b \right) - G\left( b \right) - F\left( a \right) + G\left( a \right),\>

where \(F\left( x \right)\) and also \(G\left( x \right)\) space antiderivatives that the attributes \(f\left( x \right)\) and \(g\left( x \right),\) respectively.

Note that this area will always be non-negative as \(f\left( x \right) - g\left( x \right) \ge 0\) for every \(x \in \left< a,b \right>.\)

If there room intersection points, we should break up the interval into several subintervals and also determine which curve is higher on every subinterval. Climate we have the right to determine the area of each an ar by complete the difference of the larger and also the smaller function.

Area in Polar Coordinates

Consider the an ar \(OKM\) bounded by a polar curve \(r = f\left( \theta \right)\) and also two semi-straight present \(\theta =\alpha\) and \(\theta = \beta.\)

Figure 3.

The area that the polar region is provided by


The area the a region between 2 polar curve \(r = f\left( \theta \right)\) and \(r = g\left( \theta \right)\) in the ar \(\left< \alpha ,\beta \right>\) is express by the integral

d\theta .\>
" width="800" height="779.5">Figure 4.

Area of a an ar Bounded by a Parametric Curve

Recall that the area under a curve \(y = f\left( x \right)\) for \(f\left( x \right) \ge 0\) on the expression \(\left< a,b \right>\) have the right to be computed through the integral \(\int\limits_a^b f\left( x \right)dx.\) Suppose currently that the curve is defined in parametric kind by the equations


If the parameter \(t\) runs in between \(t_1\) and also \(t_2\) where


then the area under the curve is offered by the formula


The functions \(x\left( t \right),\) \(x^\prime\left( t \right),\) \(y\left( t \right)\) here are assumed come be consistent on the expression \(\left< a,b \right>.\) besides that, the function \(x\left( t \right),\) need to be monotonic ~ above this interval.

Figure 5.

If \(x = x\left( t \right),\) \(y = y\left( t \right),\) \(0 \le t \le T\) are parametric equations the a smooth piecewise closed curve \(C\) traversed in the counterclockwise direction and bounding a an ar on the left (Figure \(5\)), climate the area the the region is given by the following integrals:

dt .\>

Solved Problems

Click or insanity a problem to check out the solution.

Example 1

Find the area between the curves \(y = x^3\) and also \(y = 3x + 2.\)

Example 2

At what worth of the parameter \(b\left( b \gt 1 \right)\) the area under the curve \(y = x^2\) ~ above the expression \(\left<1,b\right>\) is equal to \(1?\)

Example 3

Find the name: coordinates of the point \(a\) the splits the area under the root duty \(y = \sqrtx\) on the term \(\left<0,4\right>\) into equal parts.

Example 4

The region is bounded by the vertical lines \(x = t\), \(x = t + \frac\pi 2\), the \(x-\)axis, and also the curve \(y = a + \cos x,\) where \(a \ge 1.\) identify the value of \(t\) in ~ which the an ar has the largest area.

Example 1.

Find the area in between the curves \(y = x^3\) and \(y = 3x + 2.\)


First we determine the point out of intersection the the curves.

We collection \(f\left( x \right) = g\left( x \right)\) to discover the roots:

\< \Rightarrow x^3 + \underbrace x^2 - x^2_0 - 3x - 2 = 0,\>
\< \Rightarrow x^3 + x^2 - x^2 - x - 2x - 2 = 0,\>
\< \Rightarrow x^2\left( x + 1 \right) - x\left( x + 1 \right) - 2\left( x + 1 \right) = 0,\>
\< \Rightarrow \left( x + 1 \right)\left( x^2 - x - 2 \right) = 0.\>

Solve the quadratic equation:


Thus, the cubic equation has actually two roots: \(x = -1\) (of multiplicity \(2\)) and \(x=2.\)

The area us wish to calculate is shown in figure \(6\) below.

Figure 6.

We can see native the figure that the line \(y = 3x + 2\) lies above the cubic parabola \(y = x^3\) on the interval \(\left< - 1,2 \right>.\) Hence, the area that the an ar is provided by

dx = \int\limits_ - 1^2 \left( 3x + 2 - x^3 \right)dx = \left. \frac3x^22 + 2x - \fracx^44 \right|_ - 1^2 = \left( 6 + \cancel4 - \cancel4 \right) - \left( \frac32 - 2 - \frac14 \right) = \frac274.\>

Example 2.

At what value of the parameter \(b\left( b \gt 1 \right)\) the area under the curve \(y = x^2\) ~ above the term \(\left<1,b\right>\) is equal to \(1?\)


The area under the curve is provided by the integral


same to 1." width="800" height="967.1">Figure 7.

Integrating yields:

\<\int\limits_1^b x^2dx = \left. \fracx^33 \right|_1^b = \fracb^33 - \frac1^33 = \fracb^3 - 13 = 1.\>


\4 \approx 1.59\>

Example 3.

Find the coordinate of the suggest \(a\) that splits the area under the root role \(y = \sqrtx\) top top the expression \(\left<0,4\right>\) right into equal parts.


Figure 8.

Equating both areas, we acquire the following:

\16 \approx 2.52\>

Example 4.

The region is bounded by the vertical lines \(x = t\), \(x = t + \frac\pi 2\), the \(x-\)axis, and also the curve \(y = a + \cos x,\) whereby \(a \ge 1.\) recognize the worth of \(t\) in ~ which the region has the biggest area.


with the biggest area." width="800" height="626.6">Figure 9.

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The area of the an ar is created in the form


Using the distinction of sines identity

\<\sin \alpha - \sin \beta = 2\cos \frac\alpha + \beta 2\sin \frac\alpha - \beta 2,\>

we obtain


The region has the biggest area once \(\cos \left( t + \frac\pi 4 \right) = -1.\)

Solving this equation, we find

\<\cos \left( t + \frac\pi 4 \right) = - 1,\;\; \Rightarrow t + \frac\pi 4 = \pi + 2\pi n,\;\; \Rightarrow t = \frac3\pi 4 + 2\pi n,\,n \in \mathbbZ.\>

See an ext problems on web page 2.

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Alex Svirin, PhD
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