Video: Finding the General Antiderivative of a Function Using the Power Rule of Integration with Fractional Exponents

Determine ∫ 7√(π‘₯Β³) dπ‘₯.

02:29

Video Transcript

Determine the integral of seven times the square root of π‘₯ cubed with respect to π‘₯.

The question wants us to integrate seven multiplied by the square root of π‘₯ cubed with respect to π‘₯, but we don’t know how to integrate this directly. This means we’re going to need to manipulate or integrand into a form which we can integrate. Luckily, we can do this by using our laws of exponents. First, we recall that the square root of π‘Ž is the same as saying π‘Ž to the power of one-half. Using this, we can rewrite our integral as the integral of seven times π‘₯ cubed to the power of one-half with respect to π‘₯.

But we’re still not done. We still can’t integrate this directly. We need to use another one of our laws of exponents. We recall that π‘Ž to the power of 𝑏 all raised to the power of 𝑐 is equal to π‘Ž to the power of 𝑏 times 𝑐. Using this, we can show that π‘₯ cubed all raised to the power of one-half is equal to π‘₯ to the power of three over two. So we now need to integrate seven times π‘₯ to the power of three over two with respect to π‘₯. And we can do this by using the power rule for integration.

We recall the power rule for integration tells us, for constants π‘Ž and 𝑛 where 𝑛 is not equal to negative one, the integral of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž times π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝑐. We add one to our exponent of π‘₯ and then divide through by this new exponent of π‘₯. In our case, the exponent of π‘₯ is three over two, so we’ll set 𝑛 to be three over two. So applying our power rule of integration with π‘Ž equal to seven and 𝑛 equal to three over two, we get seven times π‘₯ to the power of three over two plus one divided by three over two plus one plus a constant of integration 𝑐.

And we can simplify this since three over two plus one is equal to five over two. So this gives us seven times π‘₯ to the power of five over two divided by five over two plus 𝑐. And to make this easier, instead of dividing by the fraction five over two, we’ll multiply by the reciprocal. And the reciprocal of five over two is two-fifths. This gives us seven times two-fifths times π‘₯ to the power of five over two plus 𝑐. And, finally, we can simplify seven times two-fifths to be 14 over five. Therefore, we’ve shown the integral of seven times the square root of π‘₯ cubed with respect to π‘₯ is equal to 14π‘₯ to the power of five over two divided by five plus the constant of integration 𝑐.

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