Antiderivatives, Section 5.1 - Area and Estimating w/Finite Sums and
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Section 4.8 - Antiderivatives If the following functions represent the derivative of the original function, find the original function. π π₯ = 3π₯ 2 πΉ π₯ = π₯3 π π₯ = π₯ 2 + 2π₯ πΉ π₯ = π π₯ = 3π₯ 2 − 1 πΉ π₯ = π₯3 − π₯ π π₯ =− 2 −3 + 4 = −2π₯ + 4 π₯3 1 3 π₯ + π₯2 3 πΉ π₯ = π₯ −2 + 4π₯ Antiderivative – If F’(x) = f(x) on an interval, then F(x) is the antiderivative of f(x) for every value of x on the interval. Section 4.8 - Antiderivatives State the derivative of each function. π π₯ = π₯3 π′ π₯ = 3π₯ 2 π π₯ = π₯3 + 2 π′ π₯ = 3π₯ 2 π π₯ = π₯3 − 1 π′ π₯ = 3π₯ 2 π π₯ = π₯3 + 4 π′ π₯ = 3π₯ 2 Theorem: – If F(x) is an antiderivative of f(x) on an interval I, then the general antiderivative of f(x) is: πΉ π₯ +πΆ π π₯ = 3π₯ 2 → πΉ π₯ = π₯ 3 + πΆ Section 4.8 - Antiderivatives Antiderivative Formulas where k is a constant (from page 281 of the textbook) Section 4.8 - Antiderivatives Write the general antiderivative of each of the following functions. π π₯ = π₯5 π π₯ = sin(2x) π₯ 5+1 π₯6 πΉ π₯ = +πΆ = +πΆ 5+1 6 1 πΉ π₯ = − cos 2π₯ + πΆ 2 π π₯ = π −3π₯ 1 −3π₯ πΉ π₯ =− π +πΆ 3 Section 4.8 - Antiderivatives Indefinite Integrals 6π₯ 1+1 5 − 6π₯ ππ₯ = 5π₯ − +πΆ 1+1 −5π πππ‘ ππ‘ = 2π π₯ − 3π −2π₯ 6π₯ 2 = 5π₯ − +πΆ 2 = 5π₯ − 3π₯ 2 + πΆ −5(−πππ π‘) +πΆ = 5πππ π‘ + πΆ ππ₯ = 2π π₯ 3 3π −2π₯ π₯ + π −2π₯ + πΆ +πΆ = 2π − 2 −2 Section 4.8 - Antiderivatives Initial Value Problems Solve for the original equation if π2 π¦ = 2 ππ₯ 2 − 6π₯ π2π¦ ππ₯ 2 = 2 − 6π₯ given π¦ ′ 0 = 4 and π¦ 0 = 1. ππ¦ = ππ₯ 2π₯ − 3π₯ 2 + 4 ππ¦ 6π₯ 2 = 2π₯ − +πΆ ππ₯ 2 2π₯ 2 3π₯ 3 π¦= − + 4π₯ + C 2 3 ππ¦ = 2π₯ − 3π₯ 2 + πΆ ππ₯ π¦ = π₯ 2 − π₯ 3 + 4π₯ + C 4 = 2 0 − 3(0)2 +πΆ 1 = (0)2 − 0 4=πΆ 1=πΆ ππ¦ = 2π₯ − 3π₯ 2 + 4 ππ₯ π¦ = π₯ 2 − π₯ 3 + 4π₯ + 1 3 + 4(0) + C Section 5.1 – Area and Estimating Finite Sums Estimating Area Under a Curve Approximate the area under the curve π π₯ = π₯ 2 + 1 from π₯ = 0 to π₯ = 2 using 2 rectangles. 2−0 π΄πππ = βπππβπ‘ β πππππ‘β = π(π₯) β βπ₯ βπ₯ = = 1. 2 Left-hand endpoints 1 2 Right-hand endpoints Midpoints 1 1 2 2 πΈπ π‘ππππ‘ππ π΄πππ πππππ π‘βπ πΆπ’ππ£π = π‘βπ π π’π ππ π‘βπ ππππ ππ πππ ππππ‘ππππππ π΄ = π 0 β 1 + π(1) β 1 π΄ = π 1 β 1 + π(2) β 1 π΄ = π .5 β 1 + π(1.5) β 1 π΄=1β1+2β1 π΄=2β1+5β1 π΄=3 π΄=7 π΄ = 1.25 β 1 + 3.25 β 1 π΄ = 4.5 Section 5.1 – Area and Estimating Finite Sums Estimating Area Under a Curve Approximate the area under the curve π π₯ = π₯ 2 + 1 from π₯ = 0 to π₯ = 2 using 4 rectangles. 2−0 1 π΄πππ = βπππβπ‘ β πππππ‘β = π(π₯) β βπ₯ βπ₯ = = . 4 2 Left-hand endpoints 1 2 Right-hand endpoints Midpoints 1 1 2 2 πΈπ π‘ππππ‘ππ π΄πππ πππππ π‘βπ πΆπ’ππ£π = π‘βπ π π’π ππ π‘βπ ππππ ππ πππ ππππ‘ππππππ LH π΄ = π 0 β .5 + π .5 . 5 + f 1 . 5 + π 1.5 . 5 = 3.75 RH π΄ = π .5 β .5 + π 1 . 5 + f 1.5 . 5 + π 2 . 5 = 5.75 Mid π΄ = π .25 β .5 + π .75 . 5 + f 1.25 . 5 + π 1.75 . 5 = 4.625 Section 5.1 – Area and Estimating Finite Sums Average Value of an Integral Average Value: Given a closed interval for a continuous function, the average value is the function value that when multiplied by the length of the interval produces the same area as that under the curve. ππππ π’ππππ π‘βπ ππ’ππ£π π΄π£πππππ ππππ’π (π΄π) = πππππ‘β ππ π‘βπ πππ‘πππ£ππ AV AV AV Section 5.1 – Area and Estimating Finite Sums Average Value of an Integral Estimate the average value for the function π π₯ = π₯ 2 on the interval 0,4 using four midpoint subintervals (rectangles) on equal width. π¬ππππππππ π¨πππ πΌππ ππ πππ πͺππππ π€πππ‘β ππ π π’ππππ‘πππ£πππ = βπ₯ = 4−0 4 = 1. π΄ = π .5 β 1 + π 1.5 β 1 + f 2.5 β 1 + π 3.5 β 1 π΄ = 21 1 2 3 4 ππππ πππ ππ πππ πππππ π¨ππππππ π½ππππ (π¨π½) = ππππππ ππ πππ ππππππππ π΄π£πππππ ππππ’π (π΄π) = 21 π΄π = = 5.25 4 21 4−0 Section 5.2 – Sigma Notation and Limits of Finite Sums Sigma Notation – A mathematical notation that represents the sum of many terms using a formula. Sequence – a function whose domain is positive integers. ππππππππ π΅πππππππ π π₯ , π π₯ , β(π₯) Sigma Notation ππ , ππ , ππ π π₯ = 2π₯ + 1 ππ = 2π + 1 g π₯ = π₯ 2 − 3π₯ + 7 π₯+6 β π₯ = 2 π₯ + 2π₯ + 3 ππ = π 2 − 3π + 7 π+6 ππ = 2 π + 2π + 3 Section 5.2 – Sigma Notation and Limits of Finite Sums Sigma Notation Examples π π ππ π π=π π=π 2 1 + 2 2 + 2 3 + 2(4) 2+2+2+2+2+2 12 2+4+6+8 20 π (ππ −ππ + π) π=π (ππ −π β π + π) + (ππ −π β π + π) + (ππ −π β π + π) π+π+π ππ Section 5.2 – Sigma Notation and Limits of Finite Sums Sigma Notation Express the sums in sigma notation. 1 + 2 + 3 + 4 + β― + 98 1 − 2 + 3 − 4 + β― − 98 ππ ππ (−π)π+π π π π=π 1 1 1 1 1 + + + + β―+ 2 3 4 70 ππ π=π π π π=π 1 1 1 1 1− + − + β―− 4 9 16 49 π (−π)π+π π=π π ππ Section 5.2 – Sigma Notation and Limits of Finite Sums Sigma Notation Linearity of Sigma π π πππ = π ππ π=π π π=π π (ππ ± ππ ) = π=π π ππ ± π=π ππ π=π Example π (πππ + ππ − π) π=π π π πππ + π=π ππ − π=π π π π π=π → π ππ + π π π=π π π− π=π π π=π Section 5.2 – Sigma Notation and Limits of Finite Sums Summation Rules π π=πβπ π=π π π=π π π(π + π) π= π π(π + π)(ππ + π) π = π π π=π π π=π π(π + π) π π = π π Section 5.2 – Sigma Notation and Limits of Finite Sums Summation Rules Examples ππ π= ππ β π = πππ πππ(πππ π, π, π, ππ, π ) π=π ππ π=π ππ(ππ + π) π= = πππ π ππ ππ π=π π π=π ππ(ππ + π)(π β ππ + π) = = ππππ π π(π + π) π π = π π = ππππ πππ(πππ π, π, π, ππ, π ) πππ(πππ ππ , π, π, ππ, π ) πππ(πππ ππ , π, π, π, π )
