What does a negative definite integral tell you?

The curve is below the x-axis over the interval (or the negative parts outweigh the positive).

How do you find the area of a region below the x-axis?

Take the modulus of \int y\,dx over that region.

A curve crosses the axis between the limits. What do you do?

Split the integral at the root and add the magnitudes of the parts.

How do you find the limits when the region is bounded by the x-axis?

Solve y=0 for the points where the curve meets the axis.

Evaluate \int0^2 x^2\,dx.

\left[\frac{x^3}{3}\right]0^2=\frac83.

Is \int x\,dy required for this topic?

No; only area given by \int y\,dx is required.

Why can \int0^3(x^2-4x+3)\,dx=0 even though area is non-zero?

The area above the axis cancels the equal area below it in the signed total.

Definite integrals and area under a curveEdexcel International A Level Maths: Flashcards

Card 1 of 120 of 12 known

Question

Evaluate $\int_a^b f(x)\,dx$.

Tap or press Space to reveal

Tap card or press Space to flip

See all 12 cards

Evaluate $\int_a^b f(x)\,dx$ .: $F(b)-F(a)$ , where $F$ is an antiderivative of $f$ .
Why is $+c$ not needed in a definite integral?: It cancels in $F(b)-F(a)$ .
Area between $y=f(x)\ge0$ , the $x$ -axis, $x=a$ and $x=b$ ?: $\int_a^b y\,dx$ .
What does a negative definite integral tell you?: The curve is below the $x$ -axis over the interval (or the negative parts outweigh the positive).
How do you find the area of a region below the $x$ -axis?: Take the modulus of $\int y\,dx$ over that region.
A curve crosses the axis between the limits. What do you do?: Split the integral at the root and add the magnitudes of the parts.
How do you find the limits when the region is bounded by the $x$ -axis?: Solve $y=0$ for the points where the curve meets the axis.
Integrate $x^n$ ( $n\ne-1$ ).: $\frac{x^{n+1}}{n+1}$ .
Rewrite $\sqrt{x}$ and $\frac{1}{x^2}$ before integrating.: $x^{\frac12}$ and $x^{-2}$ .
Evaluate $\int_0^2 x^2\,dx$ .: $\left[\frac{x^3}{3}\right]_0^2=\frac83$ .
Is $\int x\,dy$ required for this topic?: No; only area given by $\int y\,dx$ is required.
Why can $\int_0^3(x^2-4x+3)\,dx=0$ even though area is non-zero?: The area above the axis cancels the equal area below it in the signed total.

Exam questions on Definite integrals and area under a curve

The curve $C$ has equation $y=3x^2-2x+1$ , and $C$ lies above the $x$ -axis for all values of $x$ .
Hence find the area of the region bounded by $C$ , the $x$ -axis and the lines $x=-1$ and $x=2$ .2 marks
The curve $C$ has equation $y=x^2-4x+3$ and crosses the $x$ -axis at $A(1,0)$ and $B(3,0)$ .
Find the total area of the regions bounded by $C$ , the $x$ -axis, the $y$ -axis and the line $x=3$ .2 marks
The curve $C$ has equation $y=3\sqrt{x}+\frac{2}{x^2}$ , for $x>0$ .
Find $\int y\,dx$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).