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Year 12 Maths Advanced (2027) Integral calculus

Areas Involving Exponential & Logarithmic Functions

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Integral calculus

Areas involving exponential and logarithmic functions use definite integrals of \(e^{x}\), \(\ln x\) and related curves to find the regions they bound.

Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Integral calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

Areas under exponential and log curves use the same definite-integral method with the exponential and logarithmic rules. This Year 12 Mathematics Advanced topic (MAV-12-05) evaluates such areas.

Area \(=\displaystyle\int_a^b f(x)\,dx\). Use \(\int e^{ax}dx=\dfrac1a e^{ax}+C\) and \(\int\dfrac1x dx=\ln|x|+C\). \(e^{x}\) is always positive, so its area is just the integral.

Area under an exponentialThe area under y = e to the x from 0 to 1 is e minus 1. xy y=e^x 1
Area under \(y=e^{x}\) from \(0\) to \(1\) is \(e-1\).
\[\int e^{ax}\,dx=\dfrac1a e^{ax}+C,\qquad \int\dfrac1x\,dx=\ln|x|+C\]
integral of e to the a x is one over a e to the a x; integral of 1 over x is ln mod x

Method

  1. Choose the exponential or logarithmic rule.
  2. Evaluate the primitive at the limits.
  3. Leave exact answers such as \(e-1\).
Example 1 — Under e^x
Find the area under \(y=e^{x}\) from \(0\) to \(1\).
Solution
\(\ \)\(=\)\(\big[e^{x}\big]_0^1=e-1\)
Example 2 — Under e^{2x}
Find the area under \(y=e^{2x}\) from \(0\) to \(1\).
Solution
\(\ \)\(=\)\(\big[\tfrac12 e^{2x}\big]_0^1=\dfrac{e^{2}-1}{2}\)
Example 3 — Under 1/x
Find the area under \(y=\dfrac1x\) from \(1\) to \(e\).
Solution
\(\ \)\(=\)\(\big[\ln x\big]_1^{e}=1\)
Example 4 — Larger interval
Evaluate \(\displaystyle\int_0^2 e^{x}\,dx\).
Solution
\(\ \)\(=\)\(\big[e^{x}\big]_0^2=e^{2}-1\)

Common pitfalls

Divide by \(a\) in \(\int e^{ax}dx\).
\(\ln1=0\) makes many log areas tidy.
Leave exact answers unless a decimal is asked for.

Frequently asked questions

What is the area under e to the x from 0 to 1?

It is e minus 1, from e to the x evaluated between 0 and 1.

How do you find the area under e to the 2x?

Integrate to get one half e to the 2x, then evaluate between the limits.

What is the area under 1 over x from 1 to e?

It is ln e minus ln 1, which is 1 minus 0, so 1.

Should you round these areas?

Leave exact answers like e minus 1 unless the question asks for a decimal approximation.