Distribution function of $X$ at location $1.8$

Question

Let $X$ be a continuous random variable with density $$f_X(x)=x^2 I_{(1,1]}(x)+\left(\frac{7}{4}-\frac{3}{4}x\right)I_{(1,7/3)}(x)$$

The value of the distribution of $X$ at location $1.8$ is in:

$(A) (0.80,0.83]\\ (B)(0.77,0.80].\\ (C) (0.83,0.86].\\(D) (A)-(C) \text{ false }$

Here is my attempt to solve this:

$$F(1.8)=P(x\leq 1.8)=\int^1_0x^2+\int^{1.8}_1\left(\frac{7}{4}-\frac{3}{4}x\right)I_{(1,7/3)}(x)=\frac{x^3}{3}|^1_0+\left(\frac{7}{4}x-\frac{3}{4}\frac{x^2}{2}\right)^{1.8}_1\\=1/3+\frac{7}{4}(0.8)-\frac{3}{8}(3.24)=0.33+1.4-1.215=0.515$$

So the correct answer would be $D$. Is this correct?

Answer 1

I think you forget to subtract $1$.

$$F(1.8)=P(x\leq 1.8)=\int^1_0x^2+\int^{1.8}_1\left(\frac{7}{4}-\frac{3}{4}x\right)I_{(1,7/3)}(x)=\frac{x^3}{3}|^1_0+\left(\frac{7}{4}x-\frac{3}{4}\frac{x^2}{2}\right)^{1.8}_1\\=1/3+\frac{7}{4}(0.8)-\frac{3}{8}(3.24\color{red}{-1})\approx0.33+1.4-\color{red}{0.84}=0.89$$