Math, asked by jayretharekar, 7 hours ago

Area under the curve y=v3x + 4 between x=0 and x=4 is​

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Answered by akhansana5
0

Step-by-step explanation:

m=3x+4

dm=3dx

When x=0  ,  m=4  , 

x=4  ,  m=16

∫043x+4dx

=31∫416mdm

=31[23m23]416

92[43−23]=92×56squnits

=9112squnits

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SIMILAR QUESTIONS

The area bounded by the curve y=4−x2 and the line y=0 is

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