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# Tutor profile: Derrek S.

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Derrek S.
Math Tutor for Friends over the Years, Physics for Engineers
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## Questions

### Subject:Calculus

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Question:

Find the indefinite integral of the given equation: f(x) = xe^(4x)

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Derrek S.

To integrate this function, Integration by Parts is the best way to approach. Integration by parts is the inverse of the product rule in differential calculus. The formula is (int) udv = uv- (int) vdu Let u = x and v = e^(4x); differentiate each to find du and dv respecitvely du=dx and dv = 4e^(4x) No substitute into the formula (int) (x)(4e^(4x)) = (x)(e^(4x)) - (int) e^(4x)(dx) Now the integral on the right side of the equation is much easier to evaluate. Don't forget to move the constant on the left for the solution! (int) (4)xe^(4x) = xe^(4x)-4e^(4x) => (1/4)(xe^(4x))-e^(4x) F(x) = (1/4)(e^(4x))(x-4)

### Subject:Algebra

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Question:

Factor the expression (x^2-16)

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Derrek S.

In order to factor this expression, first, take note of how many terms there are. Being as there are only two terms, we know that the resulting factorization will be a +, - case and the constants will be the same. In order to figure out what that constant will be, take the square root of the initial constant. the square root of 16 is 4. Therefore the factored expression will be (x+4)(x-4)

### Subject:Physics

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Question:

A ball is thrown into the air at a velocity of 5 m/s from a height 100m above the ground. How long will it take hit the ground? What is the maximum height of the ball? The acceleration due to gravity is 16m/s^2.

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Derrek S.

First, list the given values. The ball starts 100m from the ground with an initial velocity of 5m/s. The earth puts a force on the ball, of -16m/s^2. The equation of the position of the ball is -16x^2+5x+100 from the information given. In order to find the time at which the ball is to hit the ground, you set the position equation equal to 0 and solve the resulting quadratic equation. The time the ball will reach the ground is approximately 2.66 seconds. In order to find the maximum height of the ball, set the derivative of the position function equal to 0 to find the time at which the maximum is reached, then plug that value into the position function to find the distance from the ground at maximum height. v(t)=-32t+5=0 => t=5/32 p(5/32)= -16(5/32)^2+5(5/32)+100 = 100.39 meters

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