Answer:
V = π /2 volume units
Step-by-step explanation:
Volume of y = √x bounded by y = x and y = 1
y = √x
limits of integration ( see Annex )
By simple inspection limits of integration are x = 0 to x = 1
or y = √x and y = x
solving these two equations
x = 0 y = 0 x = 1 y = ± 1
V = ∫ π*f(x)²*dx
V = π * ∫₀¹ x*dx = π * x²/2 |₀¹
V = π * (1/2) - 0
V = π /2 volume units
Differentiating the function
... g(x) = 5^(1+x)
we get
... g'(x) = ln(5)·5^(1+x)
Then the linear approximation near x=0 is
... y = g'(0)(x - 0) + g(0)
... y = 5·ln(5)·x + 5
With numbers filled in, this is
... y ≈ 8.047x + 5 . . . . . linear approximation to g(x)
Using this to find approximate values for 5^0.95 and 5^1.1, we can fill in x=-0.05 and x=0.1 to get
... 5^0.95 ≈ 8.047·(-0.05) +5 ≈ 4.598 . . . . approximation to 5^0.95
... 5^1.1 ≈ 8.047·0.1 +5 ≈ 5.805 . . . . approximation to 5^1.1
