Chapter 6

Integration (Pure 1)

Integration as the reverse of differentiation, definite integrals as areas, and volumes of revolution — plus the two situations where a naive integral gives the wrong area.

หลักสูตรแกนกลาง / สสวท.: แคลคูลัส (Calculus)ตรงกับ ม.6 ปริพันธ์และการประยุกต์ (เพิ่มเติม) — ปริมาตรรูปหมุนไม่มีในหลักสูตรไทยระดับนี้
Common Core (US): —
Cambridge: 9709 Pure Mathematics 1 — Integration

6.1 Indefinite integrals

The two rules Pure 1 needs

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c (n \neq = - 1) \int (a x + b)^{n} d x = \frac{( a x + b ) ^{n + 1}}{a ( n + 1 )} + c

The $\frac{1}{a}$ compensates for the chain rule that differentiation would have applied. Leaving it out is the most common error in the chapter.

ทำไมถึงเป็นแบบนั้น

Why is $n = - 1$ excluded? Because the rule would divide by zero — and the missing answer, $ln ∣ x ∣$ , belongs to a function Pure 1 has not met yet. It arrives in chapter 8.

ตัวอย่าง

A curve has $\frac{d y}{d x} = \frac{6}{2 x + 1}$ and passes through $(4, 10)$ . Find $y$ .

วิธีทำ

1.Write as $6 (2 x + 1)^{- 1/2}$
2.Integrate: $6 \cdot \frac{( 2 x + 1 ) ^{1/2}}{2 \cdot \frac{1}{2}} + c = 6 (2 x + 1)^{1/2} + c$
3.At $(4, 10)$ : $6 (3) + c = 10$

ตอบ $y = 6 2 x + 1 - 8$

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