How to find (-0.5,1) integral f(x) = ?

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2 Answers
mason m
Feb 27, 2016

int_-0.5^1f(x)dx=-2∫1−0.5f(x)dx=−2

Explanation:

Use the rule:

int_a^cf(x)dx=int_a^bf(x)dx+int_b^cf(x)dx∫caf(x)dx=∫baf(x)dx+∫cbf(x)dx

This can be split up into many parts. Just think of the aa and cc values on a continuous number line:

In this question, we can say that

int_-2^2.5f(x)dx=int_-2^-0.5f(x)dx+int_-0.5^1f(x)dx+int_1^2.5f(x)dx∫2.5−2f(x)dx=∫−0.5−2f(x)dx+∫1−0.5f(x)dx+∫2.51f(x)dx

Using the known values, this becomes

9=5+int_-0.5^1f(x)dx+69=5+∫1−0.5f(x)dx+6

Thus,

int_-0.5^1f(x)dx=-2∫1−0.5f(x)dx=−2

Jim H
Feb 28, 2016

Here is the answer to the second question. (See the comments for the first answer.)

Explanation:

int_1^-0.5 9(f(x)-5) dx = - int_-0.5^1 9(f(x)-5) dx ∫−0.519(f(x)−5)dx=−∫1−0.59(f(x)−5)dx

= -9 [int_-0.5^1 f(x) dx - int_-0.5^1 5 dx]=−9[∫1−0.5f(x)dx−∫1−0.55dx]

Now use the answer to your first question,

int_-0.5^1 f(x) dx = -2∫1−0.5f(x)dx=−2 together with the area of the rectangle

int_-0.5^1 5 dx = 5(1.5) = 7.5∫1−0.55dx=5(1.5)=7.5 to get

int_1^-0.5 9(f(x)-5) dx = -9[-2-7.5] = 85.5∫−0.519(f(x)−5)dx=−9[−2−7.5]=85.5

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