If 4∫−1f(x)dx=4 and 4∫2(3−f(x))dx=7, then which of the following is/are true:
A
4∫2f(x)dx=−1
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B
4∫2f(x)dx=1
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C
2∫−1f(x)dx=5
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D
2∫−1f(x)dx=−5
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Solution
The correct option is C2∫−1f(x)dx=5 We have 4∫2(3−f(x))dx=7⇒6−4∫2f(x)dx=7⇒4∫2f(x)dx=−1
Now, 4∫−1f(x)dx=4⇒4=2∫−1f(x)dx+4∫2f(x)dx⇒2∫−1f(x)dx−1=4⇒2∫−1f(x)dx=5