The issue is that you aren't considering the full support of cdf ofmax$\{y_1,_2,_y\}$. The full support is $(0, \infty)$. Taking a look here: https://en.wikipedia.org/wiki/Uniform_distribution_(continuous) at the definition of $F(x)$. Then consider that you'll have 1 minus this value, so for your problem you'd have: $a=200$, $b=600$ and then $1-F(y) = 1$ if $x < 200$, $1-F(y)=0$ if $x>600$ and $1-\frac{y-200}{400}$ when $y \in [200, 600]$. So the part you are missing in your calculations is:

$$\int_0^{200}dy=200.$$

which is what you're undershooting.

The portion of the integral above $600$ is all $0$ so can be safely omitted from the calculation. If you wanted to be complete, you'd write:

$$ \mathbb{E}(Y_{3:1}) = \int_0^{200}(1-F(y))dy + \int_{200}^{600}(1-F(y))dy + \int_{600}^{\infty}(1-F(y))dy $$

which is: $$ \int_0^{200}1dy + \int_{200}^{600}\left(1-\left(\frac{y-200}{400}\right)^3\right)dy + \int_{600}^{\infty}0dy, $$ which simplifies to: $$ 200 + 300 + 0. $$

The issue is that you aren't considering the full support of cdf of $Y=\text{max}\{X_1,X_2,X_3\}$. The full support is $(0, \infty)$. Taking a look here: https://en.wikipedia.org/wiki/Uniform_distribution_(continuous) at the definition of $F(x)$. Then consider that you'll have 1 minus this value, so for your problem you'd have: $a=200$, $b=600$ and then $1-F(y) = 1$ if $x < 200$, $1-F(y)=0$ if $x>600$ and $1-\frac{y-200}{400}$ when $y \in [200, 600]$. So the part you are missing in your calculations is:

$$\int_0^{200}dy=200.$$

which is what you're undershooting.

The portion of the integral above $600$ is all $0$ so can be safely omitted from the calculation. If you wanted to be complete, you'd write:

$$ \mathbb{E}(Y_{3:1}) = \int_0^{200}(1-F(y))dy + \int_{200}^{600}(1-F(y))dy + \int_{600}^{\infty}(1-F(y))dy $$

which is: $$ \int_0^{200}1dy + \int_{200}^{600}\left(1-\left(\frac{y-200}{400}\right)^3\right)dy + \int_{600}^{\infty}0dy, $$ which simplifies to: $$ 200 + 300 + 0. $$

The issue is that you aren't considering the full support of cdf ofmax$\{y_1,_2,_y\}$. The full support is $(0, \infty)$. Taking a look here: https://en.wikipedia.org/wiki/Uniform_distribution_(continuous) at the definition of $F(x)$. Then consider that you'll have 1 minus this value, so for your problem you'd have: $a=200$, $b=600$ and then $1-F(y) = 1$ if $x < 200$, $1-F(y)=0$ if $x>600$ and $1-\frac{y-200}{400}$ when $y \in [200, 600]$. So the part you are missing in your calculations is:

$$\int_0^{200}dy=200.$$

which is what you're undershooting.

The portion of the integral above $600$ is all $0$ so can be safely omitted from the calculation. If you wanted to be complete, you'd write:

$$ \mathbb{E}(Y_{3:1}) = \int_0^{200}(1-F(y))dy + \int_{200}^{600}(1-F(y))dy + \int_{600}^{\infty}(1-F(y))dy $$

which is: $$ \int_0^{200}1dy + \int_{200}^{600}(1-F(y))dy + \int_{600}^{\infty}(1-F(y))dy $$

The issue is that you aren't considering the full support of cdf ofmax$\{y_1,_2,_y\}$. The full support is $(0, \infty)$. Taking a look here: https://en.wikipedia.org/wiki/Uniform_distribution_(continuous) at the definition of $F(x)$. Then consider that you'll have 1 minus this value, so for your problem you'd have: $a=200$, $b=600$ and then $1-F(y) = 1$ if $x < 200$, $1-F(y)=0$ if $x>600$ and $1-\frac{y-200}{400}$ when $y \in [200, 600]$. So the part you are missing in your calculations is:

$$\int_0^{200}dy=200.$$

which is what you're undershooting.

The portion of the integral above $600$ is all $0$ so can be safely omitted from the calculation. If you wanted to be complete, you'd write:

$$ \mathbb{E}(Y_{3:1}) = \int_0^{200}(1-F(y))dy + \int_{200}^{600}(1-F(y))dy + \int_{600}^{\infty}(1-F(y))dy $$

which is: $$ \int_0^{200}1dy + \int_{200}^{600}\left(1-\left(\frac{y-200}{400}\right)^3\right)dy + \int_{600}^{\infty}0dy, $$ which simplifies to: $$ 200 + 300 + 0. $$