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His question is designed to be answered without a calculator. Two students wrote antiderivative formulas for k sin (kx), where k > 1. Student 1: Integral of (k sin (k x) ) d x = k squared cosine (k x) + C Student 2: Integral of (k sine (k x) ) d x = negative cosine (k x) + C Which student, if any, wrote a correct antiderivative formula? Student 1 only Student 2 only both Student 1 and Student 2 neither Student 1 nor Student 2

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ajeigbeibraheem

Answer:

Student 2 only

Step-by-step explanation:

Given that:

The written antiderivatives formulas by the students for k sin (kx), where  k > 1 are:

Student 1 : [tex]\int (k \ sin (kx)) \ dx= k^2 \ cos (kx) + C[/tex]

Student 2: [tex]\int (k \ sin (kx) ) \ dx = - cos (kx) + C[/tex]

The student that wrote the correct anti-derivate formula is Student 2 only.

This is because:

Suppose:

[tex]I = \int (k \sin (kx) ) \ dx[/tex] where; k > 1

This implies that:

[tex]I = \int (k \sin (kx) ) \ dx[/tex]

[tex]I =k \bigg ( \dfrac{-cos (kx)}{k} \bigg) + C[/tex]   because [tex]\int sin (ax) \ dx = \dfrac{-cos (ax)}{a}+C[/tex]

here;

C = constant of the integration

∴

I = - cos (kx) + C  which aligns with the answer given by student 2 only.

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