Since ![[0,4]=[0,1]\cup(1,4]](https://tex.z-dn.net/?f=%5B0%2C4%5D%3D%5B0%2C1%5D%5Ccup%281%2C4%5D)
, we can rewrite the integral as
Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:
Both integrals are quite immediate: you only need to use the power rule
to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.
Answer:
C. 7790.83 cm^3
Step-by-step explanation:
The volume of a sphere is given by
V = 4/3 pi r^3
We know the radius is 12.3
Using 3.14 for pi (This will give us an approximation, not an exact value)
V = 4/3(3.14) (12.3)^3
=7790.82984 cm^3
Answer:
Solution
verified
Verified by Toppr
m
2
−3m−1=0
m
2
−3m=1 → (1)
Third term =(
2
1
coeeficientofm)
2
(
2
1
×(−3))
2
=(
2
−3
)
2
=
4
9
Adding
4
9
to both sides of equation (1), we get
m
2
−3m+
4
9
=1+
4
9
∴ m
2
−3m+
4
9
=
4
4+9
∴ (m−
2
3
)
2
=
4
13
Taking square roots on both sides
∴ m−
2
3
=±
2
13
∴ m=
2
3
+
2
13
or m=
2
3
−
2
13
m=
2
3+
13
,
2
3−
13
are the roots of the given quadratic equation.
Answer:
(n-1)
Step-by-step explanation:
