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Solving integrals/ sqrt(3x+4)

Integral of sqrt(3x+4) dx

Limits of integration:

from to
v

The graph:

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Piecewise:

The solution

You have entered [src] 
  0               
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 |    _________   
 |  \/ 3*x + 4  dx
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4
∫403x+4 dx\int\limits_{4}^{0} \sqrt{3 x + 4}\, dx 
Integral(sqrt(3*x + 4), (x, 4, 0))
Detail solution

  1. Let u=3x+4u = 3 x + 4.

    Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

    ∫u3 du\int \frac{\sqrt{u}}{3}\, du

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫u du=∫u du3\int \sqrt{u}\, du = \frac{\int \sqrt{u}\, du}{3}

      1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫u du=2u323\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}

      So, the result is: 2u329\frac{2 u^{\frac{3}{2}}}{9}

    Now substitute uu back in:

    2(3x+4)329\frac{2 \left(3 x + 4\right)^{\frac{3}{2}}}{9}

  2. Now simplify:

    2(3x+4)329\frac{2 \left(3 x + 4\right)^{\frac{3}{2}}}{9}

  3. Add the constant of integration:

    2(3x+4)329+constant\frac{2 \left(3 x + 4\right)^{\frac{3}{2}}}{9}+ \mathrm{constant}

The answer is:

2(3x+4)329+constant\frac{2 \left(3 x + 4\right)^{\frac{3}{2}}}{9}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                   
 |                                 3/2
 |   _________          2*(3*x + 4)   
 | \/ 3*x + 4  dx = C + --------------
 |                            9       
/
∫3x+4 dx=C+2(3x+4)329\int \sqrt{3 x + 4}\, dx = C + \frac{2 \left(3 x + 4\right)^{\frac{3}{2}}}{9}
Simplify
The graph
0.04.00.51.01.52.02.53.03.5020
Plot the graph f(x)
The answer [src] 
-112/9
−1129- \frac{112}{9} 
=
= 
-112/9
−1129- \frac{112}{9} 
-112/9
Numerical answer [src] 
-12.4444444444444
-12.4444444444444

    Use the examples entering the upper and lower limits of integration.

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