باستخدام التعويض بفرض ع = 2س + 1، ما قيمة ∫ من 0 إلى 1 لـ (2س + 1)³ دس؟
- 1 10
- 2 8
- 3 12
- 4 16
This unit introduces the concept of the definite integral, building on the indefinite integral. It explains that while the indefinite integral includes an arbitrary constant, the definite integral, evaluated from a lower limit a to an upper limit b, yields a definite real number. The fundamental theorem of calculus is presented as the key tool: if f is continuous on [a, b] and F is any antiderivative, then the definite integral equals F(b) - F(a). The unit covers notation, properties (such as reversing limits, zero-width intervals, and additivity over adjacent intervals), and techniques for evaluation, including direct application of standard integrals and substitution. It also addresses integrals involving absolute values and discusses special properties for odd and even functions over symmetric intervals, stating that the integral of an odd function over [-a, a] is zero and that of an even function is twice the integral from 0 to a. Worked examples and exercises involve polynomial, trigonometric, exponential, and rational functions, with some problems requiring the use of modulus functions and substitution.
Multiple choice questions on this lesson in Pure Mathematics for 3rd secondary, First Term, with practice exercises for revision and exam preparation.
باستخدام التعويض بفرض ع = 2س + 1، ما قيمة ∫ من 0 إلى 1 لـ (2س + 1)³ دس؟
ما قيمة التكامل المحدد ∫ من 0 إلى 3 لـ |س - 1| دس؟
ما قيمة التكامل المحدد ∫ من 0 إلى 1 لـ (4س³ + 2س) دس؟
ما قيمة التكامل المحدد ∫ من -1 إلى 1 لـ (س³) دس؟
ما قيمة التكامل المحدد ∫ من 0 إلى π لـ (جا س) دس؟
إذا كان ∫ من 0 إلى 2 لـ ق(س) دس = 5 و ∫ من 2 إلى 5 لـ ق(س) دس = 3، فما قيمة ∫ من 0 إلى 5 لـ ق(س) دس؟
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