Question

Solve in attachment....​

Answer 1

Answer:

A)2

Step-by-step explanation:

we would like to integrate the following definite Integral:

$\displaystyle \int_{0} ^{1} 5x \sqrt{x} dx$

use constant integration rule which yields:

$\displaystyle 5\int_{0} ^{1} x \sqrt{x} dx$

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

$\displaystyle 5\int_{0} ^{1} x \cdot {x}^{1/2} dx$

once again use law of exponent which yields:

$\displaystyle 5\int_{0} ^{1} {x}^{ \frac{3}{2} } dx$

use exponent integration rule which yields;

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

simplify which yields:

$\displaystyle 2 {x}^{2} \sqrt{x} \bigg| _{0} ^{1}$

recall fundamental theorem:

$\displaystyle 2 ( {1}^{2}) (\sqrt{1} ) - 2( {0}^{2} )( \sqrt{0)}$

simplify:

$\displaystyle 2$

hence

our answer is A

Answer 2

Answer:

2 ( Option A )

Step-by-step explanation:

The given integral to us is ,

$\longrightarrow \displaystyle \int_0^1 5x \sqrt{x}\ dx$

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

$\longrightarrow \displaystyle I = 5 \int_0^1 x \sqrt{x}\ dx$

Now we can write √x as ,

$\longrightarrow I = \displaystyle 5 \int_0^1 x . x^{\frac{1}{2}} \ dx$

Simplify ,

$\longrightarrow I = 5 \displaystyle \int_0^1 x^{\frac{3}{2}}\ dx$

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

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

On simplifying we will get ,

$\longrightarrow \underline{\underline{ I = 2 }}$