§5.2 The Second Fundamental Theorem of Calculus

Introduction

The Second Fundamental Theorem of Calculus

Understanding Integral Functions

For $A(x) = \int_1^x(4 - 2t)\,dt$, the value $A(2)$ equals

The Second Fundamental Theorem of Calculus states that for $A(x) = \int_1^x f(t)\,dt$ with continuous $f$,

Let $F(x) = \int_0^x\frac{t}{1+t^2}\,dt$. Analysis shows $F$ increases for $x \gt 0$, decreases for $x \lt 0$, and satisfies $F = g$ where $g(x) = \frac{1}{2}\ln(1+x^2)$. The reason $F$ and $g$ coincide is

For $A(x) = \int_2^x f(t)\,dt$ with $A(2) = 0$ and subsequent signed areas $-0.5$ on $[2,3]$ and $-1$ on $[3,4]$, the value $A(4)$ equals

Suppose $A(0) = -\frac{1}{2} - \frac{\pi}{4}$ while an earlier antiderivative $F$ of the same $f$ satisfied $F(0) = -1$. The two functions satisfy

To sketch $B(x) = \int_3^x f(t)\,dt$ and $C(x) = \int_1^x f(t)\,dt$ from an existing graph of $A$, the fastest correct move is

For an integral function $F$ built from a continuous integrand $f$, the single most important structural fact is

For $F(x) = \int_0^x \frac{t}{1+t^2}\,dt$, the sign of $F'(x) = \frac{x}{1+x^2}$ shows that $F$

With $F' = f$ where $f'(x) = \frac{1-x^2}{(1+x^2)^2}$, the function $F$ is

For $A(x) = \int_2^x f(t)\,dt$ built on a quarter circle of radius $1$, we have $A(1) = -\frac{\pi}{4}$. Report this value as a decimal, rounded to three places.

For $F(x) = \int_0^x\frac{t}{1+t^2}\,dt$, midpoint computation gives $F(5) \approx 1.64038$. Report this value to three decimals.

The same accumulation satisfies $F(-10) = 2.35973$ because its integrand is odd, making $F$ even. Report $F(-10)$ to three decimals.

The function $E(x) = \int_0^x e^{-t^2}\,dt$ satisfies $E'(x) = e^{-x^2}$ for every real $x$, even though no elementary formula for $E$ exists.