Mathematics

Problem

A continuous function

h

is defined on the closed interval

[-2,4]

. It is known that

\begin{aligned} \text{(i)} & \quad \int_{-2}^{1} h(x)\,dx = 5,\\ \text{(ii)} & \quad \int_{1}^{4} h(x)\,dx = -3. \end{aligned}

The graph of

h

is entirely above the

x

-axis on

(-2,1)

and entirely below the

x

-axis on

(1,4)

. a. Part 1: (a) Describe, without performing any additional calculations, the geometric meaning of the positive value in statement (i) and the negative value in statement (ii). b. Part 2: (b) Calculate

\displaystyle \int_{-2}^{4} h(x)\,dx

and explain how its value relates to the net signed area enclosed by the graph of

h

and the

x

-axis on

[-2,4]

. State whether the total unsigned (geometric) area enclosed by the graph and the

x

-axis is greater than, equal to, or less than the absolute value of your answer, and justify your conclusion.

Intuitive explanation

The key insight is that additivity of integrals lets you combine information from different regions. Since $h$ is positive on $(-2,1)$ and negative on $(1,4)$ , each integral directly measures the signed area in that region—the first gives you the area above the axis, the second the (negative of the) area below. To find the total integral over the whole interval, just add the two given pieces together.

Intuitive explanation

Browse the problem library