A brief note on Definite Integrals . Read it before a test. Try each question at the end before you open its answer.
The Fundamental Theorem Signed area: ∫ a b f ( x ) d x \int_a^b f(x)\,dx ∫ a b f ( x ) d x is positive above the axis and negative below it. Part 1: if A ( x ) = ∫ a x f ( t ) d t A(x) = \int_a^x f(t)\,dt A ( x ) = ∫ a x f ( t ) d t and f f f is continuous, then A ′ ( x ) = f ( x ) A'(x) = f(x) A ′ ( x ) = f ( x ) . Part 2: ∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) for any antiderivative F F F . Condition: f f f must be continuous on [ a , b ] [a, b] [ a , b ] . Substitution and properties New limits: with u = g ( x ) u = g(x) u = g ( x ) , the limits become g ( a ) g(a) g ( a ) and g ( b ) g(b) g ( b ) . Split: ∫ a b f = ∫ a c f + ∫ c b f \int_a^b f = \int_a^c f + \int_c^b f ∫ a b f = ∫ a c f + ∫ c b f . Split at a sign change for ∣ f ∣ |f| ∣ f ∣ . Swap: ∫ a b f = − ∫ b a f \int_a^b f = -\int_b^a f ∫ a b f = − ∫ b a f . Even and odd: ∫ − a a f = 2 ∫ 0 a f \int_{-a}^{a} f = 2\int_0^a f ∫ − a a f = 2 ∫ 0 a f for even f f f , and 0 0 0 for odd f f f . Reflection: ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x . The limit of a sum Definition: ∫ a b f ( x ) d x = lim n → ∞ h ∑ r = 1 n f ( a + r h ) \int_a^b f(x)\,dx = \lim_{n \to \infty} h\sum_{r=1}^{n} f(a + rh) ∫ a b f ( x ) d x = lim n → ∞ h ∑ r = 1 n f ( a + r h ) with h = b − a n h = \dfrac{b - a}{n} h = n b − a . Sums: ∑ r = n ( n + 1 ) 2 \sum r = \dfrac{n(n + 1)}{2} ∑ r = 2 n ( n + 1 ) and ∑ r 2 = n ( n + 1 ) ( 2 n + 1 ) 6 \sum r^2 = \dfrac{n(n + 1)(2n + 1)}{6} ∑ r 2 = 6 n ( n + 1 ) ( 2 n + 1 ) . Reading a sum: lim 1 n ∑ r = 1 n f ( r n ) = ∫ 0 1 f ( x ) d x \lim \dfrac{1}{n}\sum_{r=1}^{n} f\left(\dfrac{r}{n}\right) = \int_0^1 f(x)\,dx lim n 1 ∑ r = 1 n f ( n r ) = ∫ 0 1 f ( x ) d x . A worked example Find ∫ 1 3 ( 3 x 2 + 2 ) d x \displaystyle\int_1^3 \left(3x^2 + 2\right)dx ∫ 1 3 ( 3 x 2 + 2 ) d x .
Antiderivative. F ( x ) = x 3 + 2 x F(x) = x^3 + 2x F ( x ) = x 3 + 2 x . Evaluate. F ( 3 ) − F ( 1 ) = ( 27 + 6 ) − ( 1 + 2 ) = 30 F(3) - F(1) = (27 + 6) - (1 + 2) = 30 F ( 3 ) − F ( 1 ) = ( 27 + 6 ) − ( 1 + 2 ) = 30 . Practice with new numbers
A program to try The program finds the area of the worked example by adding thin strips, and compares it with F ( 3 ) − F ( 1 ) F(3) - F(1) F ( 3 ) − F ( 1 ) .
Run it. Do the two numbers agree?
1. Change f to lambda x: 2 * x and F to lambda x: x ** 2. What do both lines show now?
Mistakes to avoid Forgetting F ( a ) F(a) F ( a ) . The answer is F ( b ) − F ( a ) F(b) - F(a) F ( b ) − F ( a ) . Old limits after a substitution. Change the limits with the variable. Area and signed area. Split where f f f changes sign. Wrong width in a sum. h = b − a n h = \dfrac{b - a}{n} h = n b − a . Check yourself Find ∫ 0 2 ( x + 1 ) d x \displaystyle\int_0^2 \left(x + 1\right)dx ∫ 0 2 ( x + 1 ) d x . Stuck? Show a hint Show the answerFind ∫ − 3 3 x 3 d x \displaystyle\int_{-3}^{3} x^3\,dx ∫ − 3 3 x 3 d x . Stuck? Show a hint Show the answerIf A ( x ) = ∫ 0 x sin t d t A(x) = \displaystyle\int_0^x \sin t\,dt A ( x ) = ∫ 0 x sin t d t , what is A ′ ( x ) A'(x) A ′ ( x ) ? Stuck? Show a hint Show the answerWhat happens to the limits of ∫ 0 1 x e x 2 d x \displaystyle\int_0^1 x\,e^{x^2}\,dx ∫ 0 1 x e x 2 d x when u = x 2 u = x^2 u = x 2 ? Stuck? Show a hint Show the answerWhat is the width h h h when [ 1 , 4 ] [1, 4] [ 1 , 4 ] is cut into n n n equal strips? Stuck? Show a hint Show the answerSources OpenStax, Calculus Volume 1 , Chapter 5, sections 5.2, 5.3 and 5.5. NCERT, Mathematics Part II for Class XII, Chapter 7, the sections on the definite integral, the Fundamental Theorem of Calculus and some properties of definite integrals.