Animated Solution for Mathematics - Definite Integration: The area bounded by the curves $y=f(x)$, the x-axis and the ordinates $x=1$ and $x=b$ is $(b-1)\sin (3b+4)$. Then $f(x)$ is

We start by translating our geometric reality into the language of calculus. The area under a curve $y=f(x)$ from a lower limit $a$ to an upper limit $b$ is defined by the definite integral ${\int }_a^{b}f(x)dx$.

In our case, the lower limit is fixed at $1$, and the upper limit is $b$. We can express this relationship as:

This equation is our starting point. The function $f(x)$ is currently 'trapped' inside the integral, and we must use inverse operations to set it free.

The Leibniz Rule

To differentiate an integral with a variable limit, we use the Newton-Leibniz Rule. This tool allows us to peek inside the integral and isolate the integrand.

The rule states:

Here, our lower limit $g(b)=1$ is a constant, and our upper limit $h(b)=b$. Since the derivative of a constant is zero, the derivative of the left-hand side simplifies beautifully to just $f(b)$.

The Product Rule Dance

Now, we must differentiate the right-hand side, $(b-1)\sin (3b+4)$, with respect to $b$. This requires the Product Rule: $\frac{d}{db}(u\cdot v)=u\cdot v^{\prime }+v\cdot u^{\prime }$.

Let $u=(b-1)$ and $v=\sin (3b+4)$. The derivative of $u$ is $1$, while the derivative of $v$ (using the Chain Rule) is $3\cos (3b+4)$.

Combining these components, we obtain:

Simplifying this expression, we arrive at the identity of the function in terms of $b$:

We have successfully extracted the function. Since $b$ was merely a placeholder for the variable limit, we replace $b$ with $x$ to find the general form of our function:

This result demonstrates the elegance of calculus. By applying the fundamental relationship between integration and differentiation, we have successfully solved the mystery of the trapped function.