Evaluate the following integral: ∫ 1 2 5 x 2 d x \int_1^2\!\frac{5}{x^2}\,dx ∫ 1 2 x 2 5 d x

this problem, we are asked to evaluate the following definite integral. Now I've learned up to this point that it's possible to evaluate definite integrals using the fundamental theorem of calculus part two. Now before I go ahead and use this theorem where I take the antiderivative of this function and then bound it from the two bounds on our integral, I'm first going to rewrite this inside function in a more simple way so I can do the definite integral. What this is is this is going five over x squared is gonna be the same thing as five times x to the negative two power. Recall that I can take exponents in the numerator that I see here and make them negative and bring them up to the top. Now from here, I can take this five, which is a constant, and pull it outside of the integrals. It's gonna be five times the integral from one to two of x to the negative two power, all with respect to x. Now from here I can use the power rule to find this integral. So my first process is gonna be to do the anti derivative here, which is gonna be adding one to the exponent and then dividing by that exponent. Doing this is going to give me five multiplied by x to the negative one power all divided by negative one, and that's gonna be bounded from one to two. So what I can do is rewrite this as five over x, and that whole thing's gonna be negative, bounded from one to two. The reason I can do this is because this will make the whole thing negative, and then x to the negative one, I can bring back to the denominator. So this is what we end up getting. Now from here, I need to apply my bounds, and I'm gonna do these respectively. So start with the high bound. And the high bound is going to give us negative five over two, and then what we need to do is subtract this whole thing from the low bound plugged in. So if I plug in the low bound, well, that's gonna give me my negative five over one, in which case these negative signs will cancel. So really all I'm gonna end up getting is negative five halves plus five over one, which can be rewritten and simplified to positive five over two. So five over two would be the solution to this problem, and that's how you can solve this integral. Hope you found this video helpful.

8. Definite Integrals

Fundamental Theorem of Calculus

Business Calculus

8. Definite Integrals

Fundamental Theorem of Calculus

Struggling with Business Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Evaluate the following integral:
$\int_1^2\!\frac{5}{x^2}\,dx$

$\frac53$

$\frac52$

$\frac{15}{2}$

$5$

Verified step by step guidance

Step 1: Recognize that the integral to evaluate is $ \int_1^2 \frac{5}{x^2} \, dx $. This is a definite integral, and the integrand $ \frac{5}{x^2} $ can be rewritten as $ 5x^{-2} $ for easier integration.

Step 2: Use the power rule for integration. The power rule states that $ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $ for $ n \neq -1 $. Applying this rule to $ 5x^{-2} $, the integral becomes $ 5 \cdot \frac{x^{-2+1}}{-2+1} = 5 \cdot \frac{x^{-1}}{-1} = -\frac{5}{x} $.

Step 3: Substitute the result of the indefinite integral into the definite integral formula. The definite integral $ \int_a^b f(x) \, dx $ is evaluated as $ F(b) - F(a) $, where $ F(x) $ is the antiderivative of $ f(x) $. Here, $ F(x) = -\frac{5}{x} $.

Step 4: Evaluate $ F(2) $ and $ F(1) $. Substitute $ x = 2 $ and $ x = 1 $ into $ F(x) = -\frac{5}{x} $ to find $ F(2) = -\frac{5}{2} $ and $ F(1) = -\frac{5}{1} = -5 $.

Step 5: Compute the definite integral by subtracting $ F(1) $ from $ F(2) $. This gives $ \int_1^2 \frac{5}{x^2} \, dx = F(2) - F(1) = \left(-\frac{5}{2}\right) - (-5) = -\frac{5}{2} + 5 $. Simplify this expression to find the final result.