Main content

### Course: AP®︎ Calculus AB (2017 edition) > Unit 8

Lesson 7: Definite integrals properties- Definite integrals on adjacent intervals
- Definite integral over a single point
- Switching bounds of definite integral
- Worked examples: Finding definite integrals using algebraic properties
- Using multiple properties of definite integrals
- Integrating sums of functions
- Integrating scaled version of function
- Worked examples: Definite integral properties 2
- Definite integrals over adjacent intervals
- 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
- Definite integrals properties review

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Warmup: Definite integral properties (no graph)

Apply the properties of definite integrals to evaluate definite integrals.

## Problem 1

Given ${\int}_{-5}^{-1}f(x){\textstyle \phantom{\rule{0.167em}{0ex}}}dx=4$ , ${\int}_{3}^{5}f(x){\textstyle \phantom{\rule{0.167em}{0ex}}}dx=9$ , and ${\int}_{-5}^{5}f(x){\textstyle \phantom{\rule{0.167em}{0ex}}}dx=10$ , find the following:

### Challenge

## Problem 2

Given ${\int}_{-8}^{-2}g(x){\textstyle \phantom{\rule{0.167em}{0ex}}}dx=11$ , ${\int}_{-8}^{0}g(x){\textstyle \phantom{\rule{0.167em}{0ex}}}dx=5$ , and ${\int}_{-2}^{8}g(x){\textstyle \phantom{\rule{0.167em}{0ex}}}dx=-5$ , 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)