Unit 7A

Integrals Part 1

Lesson 4: Areas Part 2

Watch the video The Definite Integral with Substitution to see additional examples of finding the new limits of integration.

As stated in the video, either method will work. 

Determine the area between the «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«/math»-axis and the line «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»y«/mi»«mo»=«/mo»«mi»x«/mi»«/math» on the interval «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»§#8722;«/mo»«mn»3«/mn»«mo»§#8804;«/mo»«mi»x«/mi»«mo»§#8804;«/mo»«mn»2«/mn»«/math».

Sketch a graph to see the bounded area.
 



To find the total area, first find the area of region «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»A«/mi»«/math», and then find the area of region «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»B«/mi»«/math». As both regions are triangles, use the formula «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»A«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»b«/mi»«mi»h«/mi»«/math».

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left center center right center left¨»«mtr»«mtd»«mi»Region«/mi»«mo»§#160;«/mo»«mi»A«/mi»«mo»:«/mo»«mo»§#160;«/mo»«mi»A«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»b«/mi»«mi»h«/mi»«/mtd»«mtd/»«mtd/»«mtd»«mi»Region«/mi»«mo»§#160;«/mo»«mi»B«/mi»«mo»:«/mo»«mo»§#160;«/mo»«mi»A«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»b«/mi»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mn»3«/mn»«/mfenced»«mfenced»«mn»3«/mn»«/mfenced»«/mtd»«mtd/»«mtd/»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mn»2«/mn»«/mfenced»«mfenced»«mn»2«/mn»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«/mtd»«mtd/»«mtd/»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«/mtd»«/mtr»«/mtable»«/math»

The total area is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»4«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«mo»+«/mo»«mn»2«/mn»«mo»=«/mo»«mn»6«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«msup»«mi»units«/mi»«mn»2«/mn»«/msup»«/math».

To evaluate the area using integrals, it seems reasonable to find «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msubsup»«mo»§#8747;«/mo»«mrow»«mo»§#8722;«/mo»«mn»3«/mn»«/mrow»«mn»2«/mn»«/msubsup»«mi»x«/mi»«mi mathvariant=¨normal¨» «/mi»«mi»d«/mi»«mi»x«/mi»«/math».

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«msubsup»«mo»§#8747;«/mo»«mrow»«mo»§#8722;«/mo»«mn»3«/mn»«/mrow»«mn»2«/mn»«/msubsup»«mi»x«/mi»«mi mathvariant=¨normal¨» «/mi»«mi»d«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msubsup»«mfenced open=¨¨ close=¨]¨»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«/mfenced»«mrow»«mo»§#8722;«/mo»«mn»3«/mn»«/mrow»«mn»2«/mn»«/msubsup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfenced»«mrow»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§#8729;«/mo»«msup»«mfenced»«mn»2«/mn»«/mfenced»«mn»2«/mn»«/msup»«/mrow»«/mfenced»«mo»§#8722;«/mo»«mfenced»«mrow»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§#8729;«/mo»«msup»«mfenced»«mrow»«mo»§#8722;«/mo»«mn»3«/mn»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#8722;«/mo»«mn»4«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«mn»2«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«/mtd»«/mtr»«/mtable»«/math»

When using geometric shapes, the area was calculated to be «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»6«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«msup»«mi»units«/mi»«mn»2«/mn»«/msup»«/math». However, when using integration, the evaluated integral was calculated to be «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»§#8722;«/mo»«mn»2«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«msup»«mi»units«/mi»«mn»2«/mn»«/msup»«/math». Why are the values different? -2.5 is called the net area (difference in area) whereas 6.5 is called the total area.

Consider looking separately at the regions above and below the «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«/math»-axis.

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left center center right center left¨»«mtr»«mtd»«mi»Above«/mi»«mo»§#160;«/mo»«mi»the«/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»x«/mi»«mo»-«/mo»«mi»axis«/mi»«mo»:«/mo»«mo»§#160;«/mo»«mi»A«/mi»«mfenced»«mn»2«/mn»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msubsup»«mo»§#8747;«/mo»«mn»0«/mn»«mn»2«/mn»«/msubsup»«mi»x«/mi»«mi mathvariant=¨normal¨» «/mi»«mi»d«/mi»«mi»x«/mi»«mo»=«/mo»«msubsup»«mfenced open=¨¨ close=¨]¨»«mrow»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfenced»«mn»0«/mn»«mn»2«/mn»«/msubsup»«/mtd»«mtd/»«mtd/»«mtd»«mi»Below«/mi»«mo»§#160;«/mo»«mi»the«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»-«/mo»«mi»a«/mi»«mi»x«/mi»«mi»i«/mi»«mi»s«/mi»«mo»:«/mo»«mo»§#160;«/mo»«mi»A«/mi»«mfenced»«mn»0«/mn»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msubsup»«mo»§#8747;«/mo»«mrow»«mo»§#8722;«/mo»«mn»3«/mn»«/mrow»«mn»0«/mn»«/msubsup»«mi»x«/mi»«mi mathvariant=¨normal¨» «/mi»«mi»d«/mi»«mi»x«/mi»«mo»=«/mo»«msubsup»«mfenced open=¨¨ close=¨]¨»«mrow»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfenced»«mrow»«mo»§#8722;«/mo»«mn»3«/mn»«/mrow»«mn»0«/mn»«/msubsup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msup»«mfenced»«mn»2«/mn»«/mfenced»«mn»2«/mn»«/msup»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msup»«mfenced»«mn»0«/mn»«/mfenced»«mn»2«/mn»«/msup»«/mtd»«mtd/»«mtd/»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msup»«mfenced»«mn»0«/mn»«/mfenced»«mn»2«/mn»«/msup»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msup»«mfenced»«mrow»«mo»§#8722;«/mo»«mn»3«/mn»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#8722;«/mo»«mn»0«/mn»«/mtd»«mtd/»«mtd/»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo»§#8722;«/mo»«mn»4«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«/mtd»«mtd/»«mtd/»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«mn»4«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«/mtd»«/mtr»«/mtable»«/math»

It seems the calculation of the area below the «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«/math»-axis yields a negative value, yet area is always positive. To adjust for this, the absolute value of the integral for that region can be used.

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»Area«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfenced open=¨|¨ close=¨|¨»«mrow»«msubsup»«mo»§#8747;«/mo»«mrow»«mo»§#8722;«/mo»«mn»3«/mn»«/mrow»«mn»0«/mn»«/msubsup»«mi»x«/mi»«mi»d«/mi»«mi»x«/mi»«/mrow»«/mfenced»«mo»+«/mo»«msubsup»«mo»§#8747;«/mo»«mn»0«/mn»«mn»2«/mn»«/msubsup»«mi»x«/mi»«mi»d«/mi»«mi»x«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»-«/mo»«msubsup»«mo»§#8747;«/mo»«mrow»«mo»-«/mo»«mn»3«/mn»«/mrow»«mn»0«/mn»«/msubsup»«mi»x«/mi»«mi»d«/mi»«mi»x«/mi»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«msubsup»«mo»§#8747;«/mo»«mn»0«/mn»«mn»2«/mn»«/msubsup»«mi»x«/mi»«mi»d«/mi»«mi»x«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfenced open=¨|¨ close=¨|¨»«mrow»«mo»§#8722;«/mo»«mn»4«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«/mrow»«/mfenced»«mo»+«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mo»-«/mo»«mo»(«/mo»«mo»-«/mo»«mn»4«/mn»«mo».«/mo»«mn»5«/mn»«mo»)«/mo»«mo»+«/mo»«mn»2«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»5«/mn»«mo»+«/mo»«mn»2«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»6«/mn»«mo».«/mo»«mn»5«/mn»«/mtd»«/mtr»«/mtable»«/math»

The value of a definite integral and the corresponding area between a curve and the «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«/math»-axis are not always the same. When evaluating a definite integral for a function whose graph lies below the «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«/math»-axis, the result will be negative. The corresponding area, however, must always be reported as a positive value.