Question

Question #64897

Answer 1

The answer is `=40/3`

Explanation:

We need

`intx^ndx=x^(n+1)/(n+1)+C(n!=-1)`

Perform the integral

`int_1^9((x-1)dx)/sqrtx=int_1^9(x/sqrtx-1/sqrtx)dx`

`=int_1^9(sqrtx-1/sqrtx)dx`

`=[2/3x^(3/2)-2sqrtx]_1^9`

`=(2/3*27-6)-(2/3-2)`

`=14-2/3`

`=40/3`

Answer 2

`10/3 or 13 1/3`

Explanation:

`int_1^9(x-1)/sqrtx`

Let us start by splitting the fraction into two parts:
`=int_1^9(x/sqrtx-1/sqrtx)`

Then we can write `1/sqrt x` in terms of power (`x^(-1/2`):
`=int_1^9((x^1*x^(-1/2))-x^(-1/2))`

Adding the powers in the first term, using following power rules:
`a^x*a^y=a^(x+y)`
`a^x/a^y=a^(x-y)`

`rarrint_1^9(x^(1-1/2)-x^(-1/2))`
`=int_1^9(x^(1/2)-x^(-1/2))`

Using the following power rule of integration,

We find out that :
`int_1^9(x^(1/2)-x^(-1/2))=[(x^(1/2+1))/(1/2+1)+C-x^(-1/2+1)/(-1/2+1)+C]_1^9`

`=[x^(3/2)/(3/2)+C-x^(1/2)/(1/2)+C]_1^9`
`=[2/3sqrt(x^3)+C-2sqrtx+C]_1^9`

Plotting the value of 9 then subtracting it from value of 1:
`=(2/3sqrt(9^3)-2sqrt9+C)-(2/3sqrt(1^3)-2sqrt1+C)`

Evaluating the function we get:
`=(2/3(3^3)-2(3)+C)-(2/3(1)-2(1)+C)`
`=(2(3^2)-6+C)-(2/3-2+C)`
`=(2(9)-6+C)-(2/3-6/3+C)`
`=12cancel+C-4/3cancel(-C)=36/3-(-4/3)=36/3+4/3`
`=40/3=13 1/3`

Answer 3

`40/3`

Explanation:

`"express the terms in the form "ax^n" and integrate"`
`"each term using the "color(blue)"power rule"`

`•color(white)(x)int(ax^n)=a/(n+1)x^(n+1)color(white)(x)n!=-1`

`int_1^9((x-1))/sqrtxdx=int_1^9((x-1))/x^(1/2)dx`

`=int_1^9(x^(1/2)-x^(-1/2))dx`

`=[2/3x^(3/2)-2x^(1/2)]_1^9`

`=(2/3xx27-6)-(2/3-2)`

`=12+4/3`

`=40/3`