Previously, we proved the first fundamental theorem of calculus (FTC). There still remains the second part of this theorem. The second FTC states that \( \int_{a}^{b}f(x)dx=F(b)-F(a) \). Since the first FTC explains that differentiating the integral gives back the function in the integrand, the integral must be equal to the antiderivative of the function \( F(x)(\frac{\mathrm{d} }{\mathrm{d} x}F(x)=f(x)) \). In other words, the indefinite integral \( \int_{}^{}f(x)=F(x)+C \) where \( C \) is an arbitrary constant. Since \( \frac{\mathrm{d} }{\mathrm{d} x}F(x)=\frac{\mathrm{d} }{\mathrm{d} x}F(x)+5=\frac{\mathrm{d} }{\mathrm{d} x}F(x)-5=f(x) \), any real constant (positive or negative) added to the antiderivative will still equal \( f(x) \) when the full function is differentiated. For this reason, the indefinite integral represents a family of functions because all real values of \( C \) satisfy the first FTC. This arbitrary constant will be important in the following proof.
While proving the first FTC, we used the function \( A(x)=\int_{a}^{x}f(t)dt \) to represent the area under \( f(x) \) from some constant \( a \) to a variable \( x \). With this, the second FTC can be represented as \( A(b) \) because \( A(b)=\int_{a}^{b}f(x)dx \) which equals \( F(b)+C \) due to the previous definition of the integral. To solve this, we must first solve \( A(a)=\int_{a}^{a}f(x)dx=F(a)+C \), however, because the width between \( a \) and \( a \) is 0, we know the area must be 0 as well so \( A(a)=0=F(a)+C \). Rearranging this, we get \( C=-F(a) \). This means that \( \int_{a}^{b}f(x)dx=F(b)-F(a) \) which is the second FTC.
