−12(7z+4)+15(5z−15) =
2 answers:
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Answer:
The area under the curve y = tsin(t-1), x = ㏑(t) from x = 0 to x = ㏑( + 1) is 2 square units
Step-by-step explanation:
y = tsin(t-1)
x = ㏑(t)
⇒dx = dt
x = 0 ⇒ t = 1 and x = ㏑(+1) ⇒ t =
+1
Area under the curve from x = 0 to x = ㏑(+1) or from t = 1 to t =
+1 is
with limits : t from 1 to
+1
=
=
Using the substitution t = k + 1 :
dt = dk
when t = 1, k = 0 and when t = +1, k =
The integral is now modified as :
= -(cos()-cos(0))
= -(-1-1)
= -(-2)
= <u>2 square units.</u>