§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

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 $\int_2^5 (3x^2 - 4x^3)\,dx$.

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 $2$.