Area in rectangular Coordinates

Recall that the area under the graph of a constant function \(f\left( x \right)\) in between the upright lines \(x = a,\) \(x = b\) deserve to be computed by the identify integral:


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

*
Figure 1.

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

We can extend the notion of the area under a curve and consider the area that the an ar between two curves.

If \(f\left( x \right)\) and also \(g\left( x \right)\) are two consistent functions and also \(f\left( x \right) \ge g\left( x \right)\) ~ above the closeup of the door interval \(\left< a,b \right>,\) then the area in between the curves \(y = f\left( x \right)\) and \(y = g\left( x \right)\) in this term is offered by


*
Figure 2.

In terms of antiderivatives, the area of region is expressed 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 \(G\left( x \right)\) are antiderivatives of the functions \(f\left( x \right)\) and \(g\left( x \right),\) respectively.

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

If there room intersection points, we need to break up the interval into several subintervals and determine i m sorry curve is greater on every subinterval. Climate we can determine the area that each region by completely the difference of the larger and also the smaller function.

Area in Polar Coordinates

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

*
Figure 3.

The area the the polar region is given by


\

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


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

Area that a region Bounded by a Parametric Curve

Recall that the area under a curve \(y = f\left( x \right)\) because that \(f\left( x \right) \ge 0\) ~ above the term \(\left< a,b \right>\) can be computed with 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 provided by the formula


\

The features \(x\left( t \right),\) \(x^\prime\left( t \right),\) \(y\left( t \right)\) below are assumed come be continuous on the expression \(\left< a,b \right>.\) besides that, the role \(x\left( t \right),\) have to be monotonic top top this interval.

*
Figure 5.

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


\
dt .\>

Solved Problems

Click or tap a difficulty to check out the solution.


Example 1

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


Example 2

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


Example 3

Find the coordinate of the point \(a\) that splits the area under the root duty \(y = \sqrtx\) top top the expression \(\left<0,4\right>\) 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 \(y = a + \cos x,\) wherein \(a \ge 1.\) identify the value of \(t\) in ~ which the region has the largest area.


Example 1.

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


Solution.

First we recognize the points of intersection the the curves.

We set \(f\left( x \right) = g\left( x \right)\) 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 we wish to calculate is presented in figure \(6\) below.

*
Figure 6.

We have the right to see from the figure that the heat \(y = 3x + 2\) lies over the cubic parabola \(y = x^3\) top top the expression \(\left< - 1,2 \right>.\) Hence, the area of the an ar is given 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 worth of the parameter \(b\left( b \gt 1 \right)\) the area under the curve \(y = x^2\) on the term \(\left<1,b\right>\) is same to \(1?\)


Solution.

The area under the curve is given by the integral


\

*
equal 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.\>

Hence


\4 \approx 1.59\>

Example 3.

Find the coordinate of the allude \(a\) the splits the area under the root duty \(y = \sqrtx\) ~ above the term \(\left<0,4\right>\) right into equal parts.


Solution.

*
Figure 8.

Equating both areas, we gain 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 \(y = a + \cos x,\) where \(a \ge 1.\) recognize the value of \(t\) in ~ which the an ar has the largest area.


Solution.

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

See more:
Got Ya A Dollar - Camille Chen State Farm Commercial

The area that the an ar is created in the form


\

Using the difference of sines identity


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

we obtain


\

The an ar has the biggest area as soon as \(\cos \left( t + \frac\pi 4 \right) = -1.\)

Solving this equation, us 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 much more problems on web page 2.


Recommended Pages


Page 1Page 2

Legal


Alex Svirin, PhD
Privacy Policy

Contact


info
rwandachamber.org

Tools


*

By utilizing our website, you agree come our cookie policy