I remember thinking about this problem in high school. Here is the intuitive explanation that popped into my head one morning before I got out of bed. I am assuming the question you are asking is why the antiderivative evaluating at the endpoints gives you the area under the curve.

Let $f(x)$ be a function on an interval $[a,b]$. We want to calculate $\int_a^b f(x)dx$, i.e. the area under the curve of $f$ from $a$ to $b$. Intuitively, $\int_a^b f(x)dx$ is the infinite sum of all rectangles with height $f(x)$ and width $dx$, and we are summing over all $x$ in the interval $[a,b]$. Let's take for granted that this expression $\int_a^b f(x)dx$ really is referring to the area under the curve.

Let $y = F(x)$ be an antiderivative of $f$. So $\frac{dy}{dx} = f$.

Choose any finite sequence of points $a = x_0 \leq \cdots \leq x_t =b$. We have the 'net change' $F(b) - F(a)$, and on the other hand we have the 'small changes' $F(x_1) - F(a), F(x_2) - F(x_1)$ etc. But if you add all the small changes:

$$[F(x_1) - F(a)] + [F(x_2) - F(x_1)] + \cdots + [F(b) - F(x_{t-1})]$$ you get the net change, $F(b) - F(a)$. The sum of the small changes is the net change. Now imagine infinitely many points there. Each small change, which I might have written as $\Delta y$, is now written as $dy$ (this is the notational tradition in calculus). And again, the net change is the (now infinite) sum of the (infinitely small) small changes: $$F(b) - F(a) = \int_a^b dy$$ But $$\int_a^b dy = \int_a^b \frac{dy}{dx}dx = \int_a^b f(x)dx$$

I remember thinking about this problem in high school. Here is the intuitive explanation that popped into my head one morning before I got out of bed. I am assuming the question you are asking is why the antiderivative evaluated at the endpoints gives you the area under the curve.

Let $f(x)$ be a function on an interval $[a,b]$. We want to calculate $\int_a^b f(x)dx$, i.e. the area under the curve of $f$ from $a$ to $b$. Intuitively, $\int_a^b f(x)dx$ is the infinite sum of all rectangles with height $f(x)$ and width $dx$, and we are summing over all $x$ in the interval $[a,b]$. Let's take for granted that this expression $\int_a^b f(x)dx$ really is referring to the area under the curve.

Let $y = F(x)$ be an antiderivative of $f$. So $\frac{dy}{dx} = f$.

Choose any finite sequence of points $a = x_0 \leq \cdots \leq x_t =b$. We have the 'net change' $F(b) - F(a)$, and on the other hand we have the 'small changes' $F(x_1) - F(a), F(x_2) - F(x_1)$ etc. But if you add all the small changes:

$$[F(x_1) - F(a)] + [F(x_2) - F(x_1)] + \cdots + [F(b) - F(x_{t-1})]$$ you get the net change, $F(b) - F(a)$. The sum of the small changes is the net change. Now imagine infinitely many points there. Each small change, which I might have written as $\Delta y$, is now written as $dy$ (this is the notational tradition in calculus). And again, the net change is the (now infinite) sum of the (infinitely small) small changes: $$F(b) - F(a) = \int_a^b dy$$ But $$\int_a^b dy = \int_a^b \frac{dy}{dx}dx = \int_a^b f(x)dx$$