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Sagot :

VirαtKσhliidn

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]

Nayefxidn

Answer:

A)2

Step-by-step explanation:

we would like to integrate the following definite Integral:

[tex] \displaystyle \int_{0} ^{1} 5x \sqrt{x} dx[/tex]

use constant integration rule which yields:

[tex] \displaystyle 5\int_{0} ^{1} x \sqrt{x} dx[/tex]

notice that we can rewrite √x using Law of exponent therefore we obtain:

[tex] \displaystyle 5\int_{0} ^{1} x \cdot {x}^{1/2} dx[/tex]

once again use law of exponent which yields:

[tex] \displaystyle 5\int_{0} ^{1} {x}^{ \frac{3}{2} } dx[/tex]

use exponent integration rule which yields;

[tex] \displaystyle 5 \left( \frac{{x}^{ \frac{3}{2} + 1 } }{ \frac{3}{2} + 1} \right) \bigg| _{0} ^{1} [/tex]

simplify which yields:

[tex] \displaystyle 2 {x}^{2} \sqrt{x} \bigg| _{0} ^{1} [/tex]

recall fundamental theorem:

[tex] \displaystyle 2 ( {1}^{2}) (\sqrt{1} ) - 2( {0}^{2} )( \sqrt{0)} [/tex]

simplify:

[tex] \displaystyle 2 [/tex]

hence

our answer is A