Many students in high school go through introductory calculus to pursue more complex fields in STEM. Quickly, students realize the importance of calculus and its concepts. Many STEM fields such as physics, chemistry, computer science, computational linguistics, biology, statistics, etc. become far more manageable with the use of calculus. Additionally, the concepts of calculus provide a basic understanding of the ideas in these various fields. Take statistics for example, the formula for the normal distribution is represented as \( f(x)=\frac{1}{\sigma\sqrt{2\pi }}e^{-\frac{1}{2}}(\frac{x-\mu }{\sigma })^{2} \) where \( \sigma \) represents the standard deviation and \( \mu \) represents the mean. The constant in front of the exponential function stems from the Gaussian integral where \( \int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi} \). With the knowledge of calculus, you can develop a deeper understanding of where such formulas come from and the reasoning behind their derivations.

Arguably, the most important idea of calculus is the relationship between differentiation and integration. Intuitively, this doesn’t make sense. How could slope and area be related? The first fundamental theorem of calculus states \( \frac{\mathrm{d} }{\mathrm{d} x}\int_{a}^{x}f(t)dt=f(x) \). In other words, to find the integral of a function, you must first solve for the antiderivative. Most students accept that differentiation and integration are inverse functions, similar to multiplication and division. However, most fail to understand why this is the case. Frankly, the proof of this relationship is quite simple.

Imagine a function \( f(x) \) where the area between the constant \( a \) and a variable \( x \) is represented as \( A(x)=\int_{a}^{x}f(t)dt \). Consider both \( A(x) \) and \( A(x+h) \) where \( h \) represents some change in \( x \). In other words, both functions represent an area that starts at the value \( a \) . The only difference is that \( A(x) \) represents the area of \( f(x) \) from \( a \) to \( x \) while \( A(x+h) \) represents the area from \( a \) to \( x+h \) of the same function. If we wanted to find the area between \( x \) and \( x+h \) we could just evaluate \( A(x+h) – A(x) \) . Since the basic idea of area under the curve is that it can be represented as the sum of infinitely many rectangles with an infinitesimally small width, we can broaden the scope of this and state that \( A(x+h)-A(x) \approx hf(x) \). Qualitatively, this equation states that this new area can be approximated with the area of a rectangle with width \( h \) and height \( f(x) \) . Using the previous concept, if \( h \) is infinitesimally small, the area of the rectangle will equal the area between \( A(x) \) and \( A(x+h) \) , so \( \lim_{h\rightarrow 0}A(x+h)-A(x)=\lim_{h\rightarrow 0}hf(x) \). Rearranging this we get \( f(x)=\lim_{h\rightarrow 0}\frac{A(x+h)-A(x)}{h} \) which is the limit definition of the derivative so \( f(x)=\frac{\mathrm{d} }{\mathrm{d} x}A(x) \). This means that the function must equal the derivative of the area function or, in other words, the area function \( A(x) \) must equal the antiderivative of \( f(x) \) . Since \( A(x)=\int_{a}^{x}f(t)dt, f(x)=\frac{\mathrm{d} }{\mathrm{d} x}A(x)=\frac{\mathrm{d} }{\mathrm{d} x}\int_{a}^{x}f(t)dt\), thus proving the first fundamental theorem of calculus.