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Solve the attachment...​

Solve The Attachment class=

Sagot :

RexRelative

Answer:

Step-by-step explanation:

=  [tex]\int\limits^1_0 {5x\sqrt{x} } \, dx[/tex]

=  [tex]\int\limits^1_0 {5xx^{1/2} } \, dx[/tex]

= [tex]\int\limits^1_0 {5x^{3/2} } \, dx[/tex]

= 5 [tex]\int\limits^1_0 {x^{3/2} } \, dx[/tex]

= 5*[tex]\frac{2}{5}[/tex]*[tex]x^{5/2}[/tex]  |[tex]\left[\begin{array}{ccc}1\\0\\\end{array}\right] \left[/tex]

= 5*[tex]\frac{2}{5}[/tex]*[tex]1^{5/2}[/tex]

= 2

VirαtKσhli

Answer:

2 ( Option A )

Step-by-step explanation:

The given integral to us is ,

[tex]\longrightarrow \displaystyle \int_0^1 5x \sqrt{x}\ dx [/tex]

Here 5 is a constant so it can come out . So that,

[tex]\longrightarrow \displaystyle I = 5 \int_0^1 x \sqrt{x}\ dx [/tex]

Now we can write √x as ,

[tex]\longrightarrow I = \displaystyle 5 \int_0^1 x . x^{\frac{1}{2}} \ dx [/tex]

Simplify ,

[tex]\longrightarrow I = 5 \displaystyle \int_0^1 x^{\frac{3}{2}}\ dx [/tex]

By Power rule , the integral of x^3/2 wrt x is , 2/5x^5/2 . Therefore ,

[tex]\longrightarrow I = 5 \bigg( \dfrac{2}{5} x^{\frac{5}{2}} \bigg] ^1_0 \bigg) [/tex]

On simplifying we will get ,

[tex]\longrightarrow \underline{\underline{ I = 2 }}[/tex]

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