§4.3 The definite integral

Introduction

The definition of the definite integral

Some properties of the definite integral

How the definite integral is connected to a function's average value

Evaluate $\int_0^1 3x\,dx$ using the geometry of the region between the line and the axis.

Evaluate $\int_{-1}^{1}\sqrt{1 - x^2}\,dx$.

Given that $\int_2^5 f(x)\,dx = 2$, the value of $\int_5^2 f(x)\,dx$ is

For $f(x) = 2x + 1$ on $[1,4]$, computations give $L_5 = 16.2$, $R_5 = 19.8$, $M_5 = 18$, and finer grids keep closing on $18$. The exact area is

A graph $g$ encloses, on consecutive unit subintervals from $-3$ to $4$, the signed areas $\frac{1}{2}$, $\frac{\pi}{4}$, $-\frac{1}{2}$, $-1$, $-\frac{1}{2}$, $\frac{\pi}{4}$, $\frac{\pi}{4}$. Then $\int_{-3}^{4} g(x)\,dx$ equals

An object moves with $v(t) = \sqrt{4 - (t-2)^2}$ on $[0,4]$, and $\int_0^4 v(t)\,dt = 2\pi$. The curve and the meaning of the integral are

For the same object, a rectangle with base the segment from $t = 0$ to $t = 4$ having the same area as the semicircle has height $\frac{\pi}{2}$, so the total distance traveled equals

Evaluate $\int_{-1}^{4}(2 - 2x)\,dx$, noting the region splits into areas $4$ above the axis and $9$ below it.

Given $\int_2^5 x^2\,dx = \frac{117}{3}$ and $\int_2^5 x^3\,dx = \frac{609}{4}$, evaluate $\int_2^5 (3x^2 - 4x^3)\,dx$.

Given $\int_0^2 g(x)\,dx = 4$ and $\int_2^5 g(x)\,dx = -1$, compute $\int_0^5 g(x)\,dx$.

Given $\int_0^5 f(x)\,dx = -1$ and $\int_0^5 g(x)\,dx = 3$, compute $\int_0^5 (f(x) + g(x))\,dx$.

Given $\int_0^5 x^3\,dx = \frac{625}{4}$ and $\int_0^5 g(x)\,dx = 3$, evaluate $\int_5^0 (2x^3 - 7g(x))\,dx$ as a decimal.

With $\int_0^4 v(t)\,dt = 2\pi$ meters over four minutes, compute the average value of $v$ on $[0,4]$, rounded to three decimals.

For $v(t) = \sqrt{4 - (t-2)^2}$, the integral $\int_0^4 v(t)\,dt = 2\pi$ because the curve is a semicircle of radius $2$.