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how to evaluate the integral
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\[\int\limits_{3}^{4} (4x-7)^{3}\] = \[\int\limits_{a}^{b} f(u) du\]
i need to find a, b, f(u) and evaluate the original integral.
did you try u = 4x - 7 ?
oh ok i got it, a=5, b=9. What would be f(u)? I put u^3 but its wrong.
u = 4x -7 differentiate both sides, what do u get?
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du = 4
do you mean du = 4dx?? or du/dx = 4
from that isolate dx
du=4dx dx=du/4
yes, if you plug that in your new integral, you'll get \(\int_5^9 u^3 \dfrac{du}{4}\) effectively making f(u) as u^3/4
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ah ok, got it. thank you.
welcome ^_^
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