IB Mathematics AHL 5.12 area of the region enclosed by a curve -AI HL Paper 1- Exam Style Questions- New Syllabus

(a)
Use substitution. Let \(u = -x^2\).
Then \(\frac{du}{dx} = -2x \Rightarrow dx = \frac{du}{-2x}\).
Substitute into the integral:
\(\int x e^u \frac{du}{-2x} = -\frac{1}{2} \int e^u du\)
\(= -\frac{1}{2} e^u + C\)
\(= -\frac{1}{2} e^{-x^2} + C\)

(b)
The area is given by the definite integral from \(0\) to \(k\).
Area \(= \int_{0}^{k} x e^{-x^2} dx\)
\(= \left[ -\frac{1}{2} e^{-x^2} \right]_{0}^{k}\)
\(= \left( -\frac{1}{2} e^{-k^2} \right) – \left( -\frac{1}{2} e^{0} \right)\)
\(= -\frac{1}{2} e^{-k^2} + \frac{1}{2}\)
\(= \frac{1}{2} (1 – e^{-k^2})\)