Indefinite
71.
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Calculus: Early Transcendentals (3rd Edition)
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- Explain the difference between an indefinite integral and a definite integral.arrow_forwardUse the differential to approximate each quantity. Then use a calculator to approximate the quantity, and give the absolute value of the difference the two results to 4decimal places. e0.002arrow_forwardUse the total differential to approximate each quantity. Then use a calculator to approximate the quantity, and give the absolute value of the differences in the two results to 4decimal places. 8.052+5.972arrow_forward
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- Evaluate the definite integral two ways: first by a u-substitution in the definite integral and then by a u-substitution in the corresponding indefinite integral. Enter the exact answer. /1 – x dx = || %3Darrow_forwardFill In The Blank Refer to the given below. Find the indefinite integral. Simplify your final answers (Just like how we simplify in our discussions). i. ii. i. iv. V. 1 Attachment(s) • i. (2x*+12x? – 5)dx S(a)(7x-5)dx ii. (-4V)dx ii. iv. (4secxtanx + csc?x)dx (sin²x+ cos²x)dx v. 4 of 4 Submitarrow_forwardEvaluate the definite integral two ways: first by a u-substitution in the definite integral and then by a u-substitution in the corresponding indefinite integral. | (4x – 4)° dx = iarrow_forward
- Evaluate the definite integral two ways: first by a u- substitution in the definite integral and then by a u- substitution in the corresponding indefinite integral. 7 (2x – 8)°dx = - iarrow_forwardArea A = 1,427 Suor Calculate the definite integral f(x) dx by referring to the Area B = 2,460 d. Area C = 3,235 figure on the right with the indicated areas. Area D = 1,677 ... f(x) dx = d. (Simplify your answer.)arrow_forwardUse substitution to find the indefinite integral. 7u =du Vu-5 Describe the most appropriate substitution case and the values of u and du. Select the correct choice below and fill in the answer boxes within your choice. O A. Substitute u for the quantity in the denominator. Let v = so that dv = () du. B. Substitute u for the quantity under the root. Let v = u- 5, so that dv = ( 1 ) du. C. Substitute u for the quantity in the numerator. Let v = that dv = () du. so Use the substitution to evaluate the integral. 7u du =arrow_forward
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