Video: Integrating Polynomials Using the Power Rule

Determine ∫(10π‘₯Β² + 21π‘₯ βˆ’ 49) dπ‘₯.

03:06

Video Transcript

Determine the integral of the function 10π‘₯ squared plus 21π‘₯ minus 49 with respect to π‘₯.

We start by recalling that the integral of the sum of two functions 𝑓 of π‘₯ and 𝑔 of π‘₯ with respect to π‘₯ is equal to the sum of the integrals of those two functions with respect to π‘₯. We can use this to split the integral in our question into the integral of 10π‘₯ squared with respect to π‘₯ plus the integral of 21π‘₯ with respect to π‘₯ plus the integral of negative 49, again, with respect to π‘₯.

Now, we can use the fact that for any constant π‘˜ the integral of π‘˜ multiplied by a function with respect to π‘₯ is equal to π‘˜ multiplied by the integral of that function with respects to π‘₯. We can use this to take the constant coefficient out of each of our three integrals. We are now ready to start evaluating these integrals. We’re going to use the power rule for integrals, which says that the integral of π‘₯ to the power of 𝑛 with respect to π‘₯ is equal to π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus our constant of integration.

In our first integral, we can see that we’re integrating π‘₯ squared. So, we have that 𝑛 is equal to two. So, by using the power rule for integrals, we have that the integral of π‘₯ squared is equal to π‘₯ to the power of two plus one all divided by two plus one plus our constant of integration 𝑐 one. In our second integral, we know that π‘₯ can be rewritten as π‘₯ to the power of one. So, we can use our power rule for integrals with 𝑛 is equal to one.

Therefore, the integral of π‘₯ with respect to π‘₯ is equal to π‘₯ to the power of one plus one divided by one plus one. And then, we add our constant of integration 𝑐 two. Finally, in our third integral, we know that π‘₯ to the power zero is equal to one. So, we can use our power rule to evaluate this integral with 𝑛 equal to zero. This gives us that the integral of one with respect to π‘₯ is equal to π‘₯ to the power of zero plus one divided by zero plus one plus our constant of integration 𝑐 three.

We can expand, simplify, and rearrange this expression to get 10π‘₯ cubed over three plus 21π‘₯ squared over two minus 49π‘₯ plus 10 multiplied by 𝑐 one plus 21 multiplied by 𝑐 two minus 49 multiplied by 𝑐 three. Since 𝑐 one, 𝑐 two, and 𝑐 three are all constants of integration, we notice that 10𝑐 one plus 21𝑐 two minus 49𝑐 three is also a constant. So, we can replace this part of the expression with a new constant which we will call 𝐢. Giving us our final answer that the integral of 10π‘₯ squared plus 21π‘₯ minus 49 with respect to π‘₯ is equal to 10π‘₯ cubed over three plus 21π‘₯ squared over two minus 49π‘₯ plus our constant of integration.

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