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Application Of Integrals

Question
CBSEENMA12035717

Using integration, find the area of the region bounded by the triangle whose vertices are (-1, 2), (1, 5) and (3, 4).

Solution

Consider the vertices, A(-1, 2), B(1, 5) and C(3, 4).
Let us find the equation of the sides of the triangle increment ABC.
Thus, the equation of AB is:
fraction numerator straight y minus 5 over denominator 5 minus 2 end fraction space equals space fraction numerator straight x minus 1 over denominator 1 plus 1 end fraction
rightwards double arrow 3 straight x minus 2 straight y plus 7 space equals space 0
Similarly comma space the space equation space of space BC space is colon
fraction numerator straight y minus 4 over denominator 4 minus 5 end fraction equals fraction numerator straight x minus 3 over denominator 3 minus 1 end fraction
rightwards double arrow straight x plus 2 straight y minus 11 space equals 0
Also comma space the space equation space of space CA space is colon
fraction numerator straight y minus 4 over denominator 4 minus 2 end fraction space equals fraction numerator straight x minus 3 over denominator 3 plus 1 end fraction
rightwards double arrow straight x minus 2 straight y plus 5 space equals 0

Now the area of increment ABC = Area of increment ADB + Area of increment BDC
therefore space Area space of space increment ADB space equals space integral subscript negative 1 end subscript superscript 1 open square brackets fraction numerator 3 straight x plus 7 over denominator 2 end fraction minus fraction numerator straight x plus 5 over denominator 2 end fraction close square brackets dx
Similarly comma space Area space of space increment BDC equals space integral subscript 1 superscript 3 open square brackets fraction numerator 11 minus straight x over denominator 2 end fraction minus fraction numerator straight x plus 5 over denominator 2 end fraction close square brackets dx
Thus comma space Area space of space increment ADB space plus space Area space of space increment BDC
equals space integral subscript negative 1 end subscript superscript 1 open square brackets fraction numerator 3 straight x plus 7 over denominator 2 end fraction minus fraction numerator straight x plus 5 over denominator 2 end fraction close square brackets dx plus integral subscript 1 superscript 3 open square brackets fraction numerator 11 minus straight x over denominator 2 end fraction minus fraction numerator straight x plus 5 over denominator 2 end fraction close square brackets dx
equals integral subscript negative 1 end subscript superscript 1 open square brackets fraction numerator 2 straight x plus 2 over denominator 2 end fraction close square brackets dx plus integral subscript 1 superscript 3 open square brackets fraction numerator 6 minus 2 straight x over denominator 2 end fraction close square brackets dx
equals integral subscript negative 1 end subscript superscript 1 open square brackets straight x plus 1 close square brackets dx plus integral subscript 1 superscript 3 open square brackets 3 minus straight x close square brackets dx
equals open square brackets straight x squared over 2 plus straight x close square brackets subscript negative 1 end subscript superscript 1 space plus space open square brackets 3 straight x minus straight x squared over 2 close square brackets subscript 1 superscript 3
equals 0 plus 2 plus 9 minus 9 over 2 minus 3 plus 1 half
equals 2 plus 9 over 2 minus 5 over 2
equals 4 space square space units