draw a rough sketch of the curve y = sqrt(3x+4) and find the area under the curve, above x-axis and - Maths - Application of Integrals

Question

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

Priyanka Kedia · Accepted Answer

$A r e a = \int_{0}^{4} y ⅆ x = \int_{0}^{4} \sqrt{3 x + 4} ⅆ x L e t 3 x + 4 = t \Rightarrow d x = \frac{1}{3} d t A n d, w h e n x = 0, t = 4 A n d, w h e n x = 4, t = 16 T h u s, A r e a = \frac{1}{3} \int_{4}^{16} \sqrt{t} ⅆ t = \frac{1}{3} \times {(\frac{{(t)}^{\frac{3}{2}}}{3 / 2})}_{4}^{16} = \frac{2}{9} \times ({(16)}^{\frac{3}{2}} - {(4)}^{\frac{3}{2}}) = \frac{2}{9} \times (64 - 8) = \frac{2}{9} \times 56 = \frac{112}{9} s q . u n i t s H e n c e, A r e a i s \frac{112}{9} s q . u n i t s$

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