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

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Answer 1 · 2016-02-27T18:48:37

#int_-0.5^1f(x)dx=-2#

Use the rule:

#int_a^cf(x)dx=int_a^bf(x)dx+int_b^cf(x)dx#

This can be split up into many parts. Just think of the #a# and #c# 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#

Using the known values, this becomes

#9=5+int_-0.5^1f(x)dx+6#

Thus,

#int_-0.5^1f(x)dx=-2#

Answer 2 · 2016-02-28T13:02:40

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

#int_1^-0.5 9(f(x)-5) dx = - int_-0.5^1 9(f(x)-5) dx #

# = -9 [int_-0.5^1 f(x) dx - int_-0.5^1 5 dx]#

Now use the answer to your first question,

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

# int_-0.5^1 5 dx = 5(1.5) = 7.5# to get

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