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Calculus, all content (2017 edition)
Course: Calculus, all content (2017 edition) > Unit 4
Lesson 6: Definite integral properties- Integrating scaled version of function
- Integrating sums of functions
- Definite integral over a single point
- Definite integrals on adjacent intervals
- Definite integral of shifted function
- Switching bounds of definite integral
- Worked examples: Definite integral properties 2
- Worked examples: Finding definite integrals using algebraic properties
- Definite integral properties (no graph): function combination
- Definite integral properties (no graph): breaking interval
- Warmup: Definite integral properties (no graph)
- Finding definite integrals using algebraic properties
- Examples leveraging integration properties
- Definite integrals properties review
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Warmup: Definite integral properties (no graph)
Apply the properties of definite integrals to evaluate definite integrals.
Problem 1
Given integral, start subscript, minus, 5, end subscript, start superscript, minus, 1, end superscript, f, left parenthesis, x, right parenthesis, d, x, equals, 4, integral, start subscript, 3, end subscript, start superscript, 5, end superscript, f, left parenthesis, x, right parenthesis, d, x, equals, 9, and integral, start subscript, minus, 5, end subscript, start superscript, 5, end superscript, f, left parenthesis, x, right parenthesis, d, x, equals, 10, find the following:
Challenge
Problem 2
Given integral, start subscript, minus, 8, end subscript, start superscript, minus, 2, end superscript, g, left parenthesis, x, right parenthesis, d, x, equals, 11, integral, start subscript, minus, 8, end subscript, start superscript, 0, end superscript, g, left parenthesis, x, right parenthesis, d, x, equals, 5, and integral, start subscript, minus, 2, end subscript, start superscript, 8, end superscript, g, left parenthesis, x, right parenthesis, d, x, equals, minus, 5, find the following:
Challenge
Want to join the conversation?
- In the final challenge question, how do we know to use the graph of y = x?(3 votes)
- The way I think about it is that a definite integral is asking for the area under the curve/graph of the function within the integral. For example, in most of the problems above, we're looking for the integral (area under the curve) of the function y=g(x). But when we need to split the integral into two in the last problem, we're left with the integral (area under the curve) of y=g(x) and the integral (area under the curve) of y=x, because x was on its own and can be considered the function by itself.(3 votes)
- Is there any way you could do this one for me? I got stuck
Integral of e^2x * (e(x)+1)^1/2
Thank you(0 votes)- = 2/15 (e x + 1)^(3/2) (3 e x - 2) + constant(3 votes)