Evaluate the definite integral of 𝑥 to the power of 88 between the limits of negative one and one.
So to enable us to evaluate our definite integral, we have this general form. And what this tells us is that if you want to find the definite integral of a function between two limits, then this is equal to the integral of the function with 𝑏, so the limit 𝑏 substituted in for 𝑥, minus the integral of the function with 𝑎, so limit 𝑎 substituted in for 𝑥. So therefore, the first stage we need to do is integrate our function 𝑥 to power of 88. So when we integrate, what we get is 𝑥 the power of 89 over 89. And then again, we’re gonna deal with the limits and the second for the definite integral, but how do we get there? Well just reminding us of how we integrate, you raise the exponent by one, so we’re gonna add one to the exponents, so it’s 𝑥 to the power of 88 plus one, so that’s where we get the 89. And then you divide by the new exponent, so the new exponent again is 88 plus one, which is 89, so that gives us 𝑥 power of 89 over 89.
So now we’re going to substitute in our values for 𝑥, and they’re gonna be our limits. And when we do that and substitute in our limits as our values of 𝑥, we’re gonna get one to the power of 89 over 89 minus then negative one to power of 89 over 89. So therefore, we’re gonna get one minus negative one over 89. That’s because it’s 89 because that was our denominator in both cases, so we’ve got common denominator. And then we have one. And we get one because one to power 89, well one to power of anything, is just one. And then minus negative one, we have negative one because we’ve got negative one to the power of 89. Well if it’s to an odd power, then it’s gonna remain negative. And now we need to remember we’re gonna have one minus negative one. Well that’s gonna be one add one because if you subtract a negative it turns into an add. So therefore, we can say that the definite integral of 𝑥 the power of 88 between the limits of negative one and one is gonna be equal to two over 89.