Question

How do you integrate `\int _ { 0} ^ { 2} ( \sqrt { x } - x + 2) d x`?

Answer 1

`int_0^2(x^(1/2)-x+2)dx~~3.89`

Explanation:

So, we are given `int_0^2(x^(1/2)-x+2)dx`

In general, `intx^ndx=x^(n+1)/(n+1)`

So, we have `int_0^2(x^(1/2)-x+2)dx=[x^(1/2+1)/(1/2+1)-x^(1+1)/(1+1)+(2x^(0+1))/(0+1)]_0^2=[x^(3/2)/(3/2)-x^2/2+2x]_0^2=((2)^(3/2)/(3/2)-(2)^2/2+2(2))-(0^(3/2)/(3/2)-0^2/2+2(0))=(2)^(3/2)/(3/2)-(2)^2/2+2(2)~~3.89`

Answer 2

`int_0^2(sqrtx-x+2)~~3.886`

Explanation:

We use the fundamental theorem of calculus, which states that:
`int_a^bf(x)dx=F(b)-F(a)` where `F(x)=intf(x)dx`

So we find the indefinite integral of `sqrtx-x+2`.
Now, remember the anti-power rule: `intx^ndx=(x^(n+1)/(n+1))` where `n!=-1`

Therefore, `int(sqrtx-x+2)dx=x^(3/2)/(3/2)-x^2/2+2x`
We simplify this: `x^(3/2)/(3/2)-x^2/2+2x=>(2x^(3/2))/3-x^2/2+2x`
`F(x)=(2x^(3/2))/3-x^2/2+2x`

We plug this into our formula: `int_0^2(sqrtx-x+2)=F(2)-F(0)`
`F(2)=(2(2)^(3/2))/3-2^2/2+2(2)` which is approximately `3.886`.

`F(0)=(2(0)^(3/2))/3-0^2/2+2(0)` which is equal to `0.`

We now subtract these two values.
`int_0^2(sqrtx-x+2)~~3.886-0`
`int_0^2(sqrtx-x+2)~~3.886`