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Question Video: Finding the General Antiderivative of a Given Function Involving Trigonometric and Root Functions Mathematics • 12th Grade

Find the most general antiderivative of the function 𝑓(π‘₯) = 4 sin π‘₯ + 3 βˆ’ (2/3√π‘₯).

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Video Transcript

Find the most general antiderivative of the function 𝑓 of π‘₯ equals four sin π‘₯ plus three minus two over three root π‘₯.

The general antiderivative of a function lowercase 𝑓 of π‘₯ is the function capital 𝐹 of π‘₯ plus 𝐢 such that the first derivative of capital 𝐹 of π‘₯ is equal to the function lowercase 𝑓 of π‘₯ and 𝐢 is any real constant. An antiderivative is not unique, and there are many functions which differ up to the value of this constant which have the same derivative. The function 𝑓 of π‘₯ here is the sum of a trigonometric term, a constant, and a power of π‘₯. Let’s simplify this final term first.

We recall that the square root function is equivalent to a power or exponent of one-half. So, this is equal to negative two over three π‘₯ to the power of one-half. Next, we recall that reciprocals are defined by negative exponents. And so, one over π‘₯ the power of a half is equal to π‘₯ the power of negative a half. And so, this final term can be rewritten as negative two-thirds π‘₯ to the power of negative one-half. So, we’ve rewritten the function 𝑓 of π‘₯ slightly.

Now, an antiderivative is linear, so the antiderivative of a sum is the sum of the antiderivatives. The process of finding an antiderivative is the reverse process of differentiating a function. And in fact, we can recall some general rules for finding antiderivatives. The most general antiderivative of the function sin of π‘Žπ‘₯, provided π‘Ž is not equal to zero, is negative one over π‘Ž multiplied by cos of π‘Žπ‘₯ plus 𝐢. This follows directly from the fact that the derivative of negative one over π‘Ž cos π‘Žπ‘₯ plus 𝐢 is sin of π‘Žπ‘₯.

Here, the argument of the sine function is simply π‘₯. So, the value of π‘Ž is one. We also know that if a derivative is multiplied by a constant, the antiderivative is multiplied by that same constant. It follows then that the antiderivative of the first term is negative four cos π‘₯ plus a constant of antidifferentiation 𝐢 one.

Now, the second term in 𝑓 of π‘₯ is just a constant, but we can think of it as three multiplied by π‘₯ to the zeroth power if we wish because π‘₯ to the zeroth power is equal to one. We can then find the antiderivatives of the second and third terms in the same way. Recalling the power rule of differentiation, we know that the first derivative of π‘₯ to the power of π‘Ž plus one over π‘Ž plus one is equal to π‘₯ to the power of π‘Ž, provided π‘Ž is not equal to negative one. So, it follows that the general antiderivative of π‘₯ to the π‘Žth power is π‘₯ the power of π‘Ž plus one over π‘Ž plus one plus 𝐢.

In practical terms, this means that to find an antiderivative of a power of π‘₯ where the power is not equal to negative one, we increase the power or exponent by one and then divide by the new exponent while also adding a constant of antidifferentiation 𝐢. So, the antiderivative of three π‘₯ to the zeroth power is three π‘₯ to the first power over one plus a constant of antidifferentiation 𝐢 two. But of course, we can just write this as three π‘₯ plus 𝐢 two.

For the third term, the antiderivative of negative two-thirds π‘₯ to the power of negative a half is negative two-thirds multiplied by π‘₯ to the power of a half. And then, we divide by a half, which is equivalent to multiplying by two plus a constant of antidifferentiation 𝐢 three. This term simplifies to negative four-thirds multiplied by π‘₯ to the power of a half. And then, we can combine the three constants of antidifferentiation into a single constant 𝐢.

Finally, we can write the terms in the antiderivative in any order, and we’ll also rewrite the exponent of a half using square root notation. And so, we have our final answer. The most general antiderivative of the function 𝑓 of π‘₯ equals four sin π‘₯ plus three minus two over three root π‘₯ is the function capital 𝐹 of π‘₯ which is equal to negative four root π‘₯ over three plus three π‘₯ minus four cos π‘₯ plus 𝐢.

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