What is $f (x) = \int \sqrt{x + 3} d x$ $f(x) = int sqrt(x+3) dx$ if $f (1) = 7$ $f(1)=7$ ?

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Answer 1 · 2018-05-26T13:52:27

$f (x) = \frac{2}{3} {(1 + 3)}^{\frac{3}{2}} + \frac{5}{3}$ $f(x)=2/3(1+3)^(3/2)+5/3$

$f (x) = \int \sqrt{x + 3} d x$ $f(x)=intsqrt(x+3)dx$

$f (x) = \int {(x + 3)}^{\frac{1}{2}} d x$ $f(x)=int(x+3)^(1/2)dx$

now

$\frac{d}{d x} {(x + 3)}^{\frac{3}{2}} = \frac{3}{2} {(x + 3)}^{\frac{1}{2}}$ $d/(dx)(x+3)^(3/2)=3/2(x+3)^(1/2)$

$f (x) = \int {(x + 3)}^{\frac{1}{2}} d x = \frac{2}{3} {(x + 3)}^{\frac{3}{2}} + c$ $f(x)=int(x+3)^(1/2)dx=2/3(x+3)^(3/2)+c$
now

$f (1) = 7$ $f(1)=7$

$:.2/3(1+3)^(3/2)+c=7$

$2/3xx8+c=7$

$16/3+c=7$

$=>c=7-16/3$

$c=5/3$

$:.f(x)=2/3(1+3)^(3/2)+5/3$

What is f(x) = int sqrt(x+3) dxf(x)=∫√x+3dxf(x) = int sqrt(x+3) dx if f(1)=7 f(1)=7f(1)=7 ?