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Determine the value of \(a\) that makes \(F(x)\) an antiderivative of \(f(x)\).

Given:
\[ f(x) = 42x^5 \]
\[ F(x) = ax^6 \]

Possible values for \(a\):
- 35
- 42
- 7
- 8

Answer :

The value of 'a' that makes F(x) an antiderivative of f(x) = 42x⁵ is a = 7.

To determine the value of 'a' that makes F(x) an antiderivative of f(x), we need to find the derivative of F(x) and check if it matches f(x).

Given that f(x) = 42x⁵ and F(x) = ax⁶ + 35x⁴ + 42x⁷, we can find the derivative of F(x) by differentiating each term separately.

The derivative of ax⁶ is 6ax⁵ (using the power rule).


The derivative of 35x⁴ is 140x³ (using the power rule).


The derivative of 42x⁷ is 294x⁶ (using the power rule).

Now, let's compare the derivative of F(x) with f(x) = 42x⁵.

6ax⁵+ 140x³ + 294x⁶ = 42x⁵

To match the terms on both sides of the equation, we can equate the corresponding coefficients.

6a = 42 (for the x⁵ term)
140 = 0 (for the x³ term)
294 = 0 (for the x⁶ term)

Simplifying the equation, we find:

6a = 42

Dividing both sides by 6:

a = 7

Therefore, the value of 'a' that makes F(x) an antiderivative of f(x) = 42x⁵ is a = 7.

In summary, to make F(x) an antiderivative of f(x) = 42x⁵, the value of 'a' needs to be 7.

To know more about antiderivative refer here:

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