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See how Nova Maths teaches Area Under a Curve — one of the most tested HSC calculus topics. Read the explanation, follow the worked example, and preview practice questions before subscribing.

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Year 12 Mathematics Advanced

Calculus

Area Under a Curve

LearnGuided practiceIndependent practiceMastery quiz

Learn how to calculate area under a curve using definite integrals.

Identify the bounds of the required area.
Set up a definite integral for area under a curve.
Find and evaluate an antiderivative using the bounds.
Check whether the curve is above or below the x-axis on the interval.
State area using square units.

Learn

Key ideas

Area under a curve can be found using a definite integral when the curve is above the x-axis on the interval.

The bounds define the interval of the area. For example, from x=a to x=b means the integral has lower bound a and upper bound b.

If the curve is below the x-axis, the definite integral is negative but the geometric area is positive.

Before giving an area answer, check whether the curve is above or below the x-axis. If the curve crosses the x-axis, split the interval as in the signed area lesson.

Area is measured in square units.

\text{Area}=\int_a^b f(x)\,dx \quad \text{if } f(x)\ge 0 \text{ on }[a,b]

\text{Area}=\left|\int_a^b f(x)\,dx\right| \quad \text{if } f(x)\le 0 \text{ on }[a,b]

\int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)

\text{area units}=\text{square units}

Worked example

Worked example 1: Area above the x-axis

\text{Find the area under }y=x^2+1\text{ from }x=0\text{ to }x=2.

Step 1: The curve is above the x-axis, so use the definite integral directly.

\text{Area}=\int_0^2 (x^2+1)\,dx

Step 2: Find an antiderivative.

\int (x^2+1)\,dx=\frac{x^3}{3}+x

Step 3: Evaluate between 0 and 2.

\left[\frac{x^3}{3}+x\right]_0^2=\frac{8}{3}+2=\frac{14}{3}

Final answer

\frac{14}{3}\text{ square units}

Guided practice preview

Try these questions

These questions are from the guided practice section. Subscribe to attempt them, check your answers and save your progress.

Question 1

Identify the lower bound:

\text{Area from }x=1\text{ to }x=4

Hint: The lower bound is the starting x-value.

Question 2

Choose the correct setup:

\text{Area under }y=x+1\text{ from }x=0\text{ to }x=2

\int_0^2 (x+1)\,dx

\int_2^0 (x+1)\,dx

\int_0^2 (x-1)\,dx

Hint: Use the function and the given bounds.

Question 3

Find an antiderivative:

\int 2x\,dx

Hint: For a definite integral, the +C cancels.

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