Solve the definite integral?

[noname1]

The problem is: ∫ x / sqrt(x²+1)

Upper limit is sqrt 3
lower limit is 0

I choose u as x²+1

1/2du = dx.x

1/2∫ 1/u^1/2 =

1/2∫ u^(-1/2)du =

1/2 ∫ U^(1/2) / (1/2) =

(1/2) * (2/1) * u^1/2 = 1 * u^(1/2)u = (sqrt3)² + 1 = 4
u = 0² + 1 = 1

than i substituted

1 * 4^1/2 - 1 * 1^(1/2) =
1 * 2 - 1 * 1 =
2 - 1 = 1

is this correct?

 

[Mark44]

The problem is: ∫ x / sqrt(x²+1)

Upper limit is sqrt 3
lower limit is 0

I choose u as x²+1

1/2du = dx.x

1/2∫ 1/u^1/2 =

1/2∫ u^(-1/2)du =

1/2 ∫ U^(1/2) / (1/2) =

(1/2) * (2/1) * u^1/2 = 1 * u^(1/2)


u = (sqrt3)² + 1 = 4
u = 0² + 1 = 1

than i substituted

1 * 4^1/2 - 1 * 1^(1/2) =
1 * 2 - 1 * 1 =
2 - 1 = 1

is this correct?

Yes. Here's another way that changes the limits of integration for the same substitution.

$$\int_0^{\sqrt{3}} \frac{x~dx}{\sqrt{x^2 + 1}}$$
$$= \frac{1}{2}\int_1^4 u^{-1/2}~du~=~u^{1/2}|_1^4$$
$$=2 - 1 = 1$$