0∫a (4-2x)dx=4 , find the value of a
Hmm. You might want to check your analysis Misaki. I get the following:
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First solve the indefinite integral
\(\int 4-2x dx=4x-x^2+c\)
Find boundaries
\(\int_{0}^{a}4-2xd=-(a-4)a-0\)
Simplify
\(-(a-4)a\)
\(-(a-4)a=4 \)Solve for a
\(a^2=8\)
\(a=\sqrt{8}\)
\(a=2\sqrt{2}\)
Ahh thank you very much!
Sorry for the mistake the correct answer is actually 2, i made a mistake during the ''solve for a'' part. So instead of \(2\sqrt{2}\) the correct answer is \(2\)
+333 
