Question Video: Integrating Polynomials Using the Power Rule | Nagwa Question Video: Integrating Polynomials Using the Power Rule | Nagwa

# Question Video: Integrating Polynomials Using the Power Rule Mathematics • Second Year of Secondary School

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Determine β«(10π₯Β² + 21π₯ β 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 π of π₯ and π 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. Weβre going to use the power rule for integrals, which says that the integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one divided by π plus one plus our constant of integration.

In our first integral, we can see that weβre integrating π₯ squared. So, we have that π 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 π 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 π 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 π 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 π 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 π 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 π one plus 21 multiplied by π two minus 49 multiplied by π three. Since π one, π two, and π three are all constants of integration, we notice that 10π one plus 21π two minus 49π three is also a constant. So, we can replace this part of the expression with a new constant which we will call πΆ. 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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