Evaluate the definite and integral.?

Question

1. x:[0,2] dx / (2x+5)
2. x:[1, e^6] dx / xsqrt(lnx)
3. x:[8,21] (e^x^(1/3)) / x^(2/3)

Answer 1

1)
∫ dx/(2x + 5)

1/2∫ 2 dx/(2x+ 5)

1/2(ln(2x + 5)

definite integral x : (0,2)

1/2[ln(4 + 5) - ln(0+5)]

1/2[ln(9) - ln(5)

1/2[ln(9/5)]

2)
∫ dx/ x sqrt(lnx)

let lnx = u

1/x dx = du

substituting

∫du/sqrt(u)

(sqrt(u)/(1/2)

(2 sqrt(u))

substituting back u = ln x

2( sqrt(ln x)

definite integral x:[1, e^6]

2[sqrt(ln e^6) - sqrt(ln 1)]

2[sqrt(6)

3)
∫ e^x^(1/3)/x^(2/3)

let x^(1/3) = u

1/3(x^(-2/3) ) dx = du

1/3(1/x^(2/3) dx = du

1/x^(2/3) = 3 du

substituting

∫ 3 e^u du

3(e^u

substituting back u = x^(1/3)

3(e^x^(1/3)

definite integral x: [8, 27]

3[e^27^(1/3) - e^8^((1/3)

3[e^3) - e^2]

Answer 2

1. x:[0,2] dx / (2x+5)

let u = 2x + 5, du = 2 dx, so dx = du/2.

rewrite in terms of u, int (du/2)/u or
(1/2) int (du/u) = .5 ln |u|, back to x
= .5 ln |2x+5| + C, with x from 0 to 2,

= .5 [ ln (2*2+5) - ln(2*0+5)]
= .5 [ ln (9) - ln(5)]
= .5 ln(9/5).


2. x:[1, e^6] dx / xsqrt(lnx) . This might need intergration by parts.

3. x:[8,21] (e^x^(1/3)) / x^(2/3)

let u = x^(1/3) du = (1/3) x^ (-2/3) dx
= (1/3) dx/ x^(2/3), or
3du = dx/x^(2/3), rewrite in terms of u

int e^u* 3du = 3 int(e^u du) = 3 e^u
= 3 e^(x^(1/3)) + C, with values
= 3 [ e^(21^1/3) - e^(8^1/3) ]
= 3 [ e^(21^1/3) - e^2 ]. Almost done, do the rest.

Answer 3

If you're this far along, you gotta know how to do integrals.
A definite integral is just the difference between the values of the integrated function evaluated at its endpoints.
x:[a,b] f'(x) dx = f(b) - f(a) when d/dc f(x) = f'(x)

Doug

Answer 4

-(1/2)*ln(5)+ln(3)

2*sqrt(6)

-3*exp(2)+3*exp(21^(1/3))