Question

Draw a rough sketch of the curve y =sqrt(3x +4) and find the area under the curve, above the x-axis and between x = 0 and x = 4.
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Answer 1

Given : curve y =sqrt(3x +4)

To Find :  the area under the curve, above the x-axis and between x = 0 and x = 4.

Solution:

y = √(3x + 4 )

x             y = √(3x + 4 )

-4/3         0

-1             1

0             2

1               2.65

2              3.16

3               3.6

4              4

y  =  √(3x + 4 )

integrate

A  =  $\int\limits^4_0 {\sqrt{3x+4}} \, dx$

=> A  =   $\left] \dfrac{(3x+4)^\dfrac{3}{2}}{3\times \dfrac{3}{2}}\right]^4_0$

= 128/9  - 16 /9

= 112/9  sq units

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