1)Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative. Remember to use ln |u| where appropriate.)
f(x) = (1/5)−(3/x)
-----> (x/5)-3lnx+C
2)Find f.
f ''(x) = 4 + 6x + 36x^2, f(0) = 2, f (1) = 10
----> 2x^2+x^3+3x^4+8x+2
A particle is moving with the given data. Find the position of the particle.
v(t) = 1.5(square root of(t)), s(4) = 17
----> s(t)=t^(3/2)+1
first: correct, also x/5)- ln(x^3)+C
second 2x^2+x^3+3x^4+8x+Cx + K
f(o)=2=k or K=2
f(1)=10=2+1+3+8+C +2 means C=10-16=-6
>>>f(x)=2x^2+x^3+3x^4+8x -6x + 2
f'= 4x + 3x^2 +12x^3 + 2
f"= 4+6x + 36x^2
how about the third one!
v = 3/2 √t
s = t^(3/2)+C
8+C = 17, so C=9
s(t) = t^(3/2) + 9
How did you get C=1?
To find the most general antiderivative of the function f(x) = (1/5) - (3/x), you can use the power rule for integration and the logarithmic rule for integrating fractions. Here's the step-by-step process:
1) Notice that the function is a sum of two terms: (1/5) and -(3/x). Each term needs to be integrated separately.
2) For the first term, (1/5), the antiderivative is simply (1/5) times x, which results in (x/5).
3) For the second term, -(3/x), you can rewrite it as -3x^(-1) to make it easier to integrate. Using the power rule, you increase the power of x by 1 (to -1 + 1 = 0) and divide by the new power (which is 0) resulting in -3 * (1/0) = -3 * ln|x|. Replace the absolute value bars with parentheses since we are dealing with a fraction. So, the antiderivative of -(3/x) is -3ln(x).
4) Combine the two antiderivatives to get the most general antiderivative: (x/5) - 3ln(x) + C, where C is the constant of integration.
To check if your answer is correct, differentiate the most general antiderivative with respect to x using the rules of differentiation. The derivative of (x/5) is (1/5), and the derivative of -3ln(x) is -(3/x).
Therefore, the derivative of (x/5) - 3ln(x) + C is (1/5) - (3/x), which is the original function f(x).
Hence, the most general antiderivative of f(x) = (1/5) - (3/x) is (x/5) - 3ln(x) + C.
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To find the function f(x) given f''(x) = 4 + 6x + 36x^2, and the initial conditions f(0) = 2 and f(1) = 10, you need to integrate twice.
1) Start by integrating f''(x) once to find f'(x). The integral of 4 is 4x, and the integral of 6x is 3x^2, and the integral of 36x^2 is 12x^3. Therefore, f'(x) = 4x + 3x^2 + 12x^3 + C1, where C1 is the constant of integration.
2) Now, integrate f'(x) to find f(x). The integral of 4x is 2x^2, and the integral of 3x^2 is x^3, and the integral of 12x^3 is 3x^4. Therefore, f(x) = 2x^2 + x^3 + 3x^4 + C1x + C2, where C1 and C2 are constants of integration.
3) To find the values of C1 and C2, use the initial conditions f(0) = 2 and f(1) = 10.
When x = 0, f(x) = 2. Plugging this into the equation, you get 2 = 0 + 0 + 0 + C1(0) + C2. This simplifies to C2 = 2.
When x = 1, f(x) = 10. Plugging this into the equation, you get 10 = 2 + 1 + 3 + C1 + 2. This simplifies to C1 = 2.
Therefore, the function f(x) = 2x^2 + x^3 + 3x^4 + 2x + 2 satisfies the given conditions f''(x) = 4 + 6x + 36x^2, f(0) = 2, and f(1) = 10.
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To find the position of a particle given the velocity function v(t) = 1.5√t and the initial condition s(4) = 17, you need to find the antiderivative of v(t).
1) Start by integrating the velocity function v(t) = 1.5√t. The integral of t^n is (2n+1)/(n+1) * t^(n+1). In this case, t^(1/2) is equivalent to the square root of t.
So, integrating v(t) = 1.5√t, you get s(t) = (2/3) * t^(3/2) + C, where C is the constant of integration.
2) Use the initial condition s(4) = 17 to find the value of C.
When t = 4, s(t) = (2/3) * 4^(3/2) + C. Simplifying this equation, you get 17 = (2/3) * 8 + C. Solving for C, you get C = 17 - (16/3) = 35/3.
3) Substitute the value of C back into the equation to find the position function. So, s(t) = (2/3) * t^(3/2) + 35/3.
Therefore, the position function of the particle is s(t) = t^(3/2) + 35/3, where t represents time.