-->

Vector Algebra

Question
CBSEENMA12032744

Using integration, find the area of the region bounded by the triangle whose vertices are (1, 0), (2, 2) and (3, 1).

Solution
Let A (1, 0), B (2, 2), C (3, 1) be the vertices of the given triangle. The equation of AB is
                                   straight y minus 0 space equals space fraction numerator 2 minus 0 over denominator 2 minus 1 end fraction left parenthesis straight x minus 1 right parenthesis comma space space space space or space space straight y space equals space 2 left parenthesis straight x minus 1 right parenthesis space space or space space straight y space equals space 2 straight x minus 2
The equation of BC is  straight y minus 2 space equals space fraction numerator 1 minus 2 over denominator 3 minus 2 end fraction left parenthesis straight x minus 2 right parenthesis comma space space or space straight y space minus 2 space equals space minus left parenthesis straight x minus 2 right parenthesis
or    straight y space equals negative straight x plus 4
Consider straight f left parenthesis straight x right parenthesis space equals space open curly brackets table attributes columnalign left end attributes row cell 2 straight x minus 2 space space if space space 1 less or equal than straight x less or equal than 2 end cell row cell 4 minus straight x space space if space space 2 less or equal than straight x less or equal than 3 end cell end table close

Let D be the foot of perpendicular from C (3, 1) on x-axis. Therefore, D is (3, 0).
   Area of quad. ABCD = integral subscript 1 superscript 3 straight f left parenthesis straight x right parenthesis space dx
                     equals space integral subscript 1 superscript 2 straight f left parenthesis straight x right parenthesis space dx space plus space integral subscript 2 superscript 3 straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 1 superscript 2 left parenthesis 2 straight x minus 2 right parenthesis space dx plus integral subscript 2 superscript 3 left parenthesis 4 minus straight x right parenthesis space dx
                       equals space open square brackets straight x squared minus 2 straight x close square brackets subscript 1 superscript 2 plus open square brackets 4 straight x minus straight x squared over 2 close square brackets subscript 2 superscript 3 space equals space open square brackets left parenthesis 4 minus 4 right parenthesis minus space left parenthesis 1 minus 2 right parenthesis close square brackets plus open square brackets open parentheses 12 minus 9 over 2 close parentheses minus left parenthesis 8 minus 2 right parenthesis close square brackets
equals space 0 plus 1 plus 15 over 2 minus 6 space equals space 5 over 2
The equation of AC is straight y minus 0 space equals space fraction numerator 1 minus 0 over denominator 3 minus 1 end fraction left parenthesis straight x minus 1 right parenthesis
or     straight y space equals 1 half left parenthesis straight x minus 1 right parenthesis
Area of increment ACD space equals space integral subscript 1 superscript 3 1 half left parenthesis straight x minus 1 right parenthesis space dx space equals space 1 half open square brackets straight x squared over 2 minus straight x close square brackets subscript 1 superscript 3
                      equals space 1 half open square brackets open parentheses 9 over 2 minus 3 close parentheses minus open parentheses 1 half minus 1 close parentheses close square brackets space equals 1 half open square brackets 3 over 2 plus 1 half close square brackets space equals space 1
therefore space space space space area space of space increment ABC space equals space Area space of space quad. space ABCD space minus space area space of space increment space ACD
                                   equals space 5 over 2 minus 1 space equals space 3 over 2 sq. space units.

Some More Questions From Vector Algebra Chapter