What is the answer of the question?

[tex] \int_0^1 \: 5x \sqrt x \: dx \\ \\ = \int_0^1 \: 5x . \: {x}^{ \frac{1}{2} } \: dx \\ \\ = 5\int_0^1 \: {x}^{ \frac{3}{2} } \: dx \\ \\ = 5 \bigg( \frac{ {x}^{ \frac{3}{2} + 1 } }{ \frac{3}{2} + 1 } \bigg)_0^1\\ \\ = 5 \bigg( \frac{ {x}^{ \frac{5}{2} } }{ \frac{5}{2}} \bigg)_0^1 \\ \\ = 5 \times \frac{2}{5} \bigg({x}^{ \frac{5}{2} } \bigg)_0^1 \\ \\ = 2 \bigg( {1}^{ \frac{5}{2} } - {0}^{ \frac{5}{2} } \bigg) \\ \\ = 2(1 - 0) \\ \\ = 2[/tex]

VirαtKσhliid VirαtKσhliid · Answer 2

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]