Question

Integration

Answer 1

\[Given~\int\limits_{k}^{2}(3-4x)dx=3,find~the~value~of~k\]

Answer 2

This is my working...\[\left[ 3x-2x^2 \right]_{k}^{2}=3\]\[\left[ 3(2)-2(2)^2)-(3(k)-2(k)^2 \right]=3\]\[2k^2-3k-5=0\]\[k=\frac{ 5 }{ 2 },k=-1\]

Answer 3

yea correct

Answer 4

But the book's answer say is k=-1

Answer 5

can anyone explain why is it k=-1 not k=-1 and k=5/2 ?

Answer 6

is it given that the solutions must be in integer form?

Answer 7

5/2 is the result of a > b ...

by convention, the upper limit higher than the lower limit. and if they are switched about we get

\[\int_{b}^{a}=-\int_{a}^{b}\]