Question

Now from the given ∫ 1 5 ​ f(3x)dx=7. Let 3x=u⇒3dx=du⇒dx= 3 du ​ , Also if x=1⇒u=3×1=3, if x=5⇒u−3×5=15 So the given integral becomes ∫ 3 15 ​ f(u) 3 du ​ =7⇒ 3 1 ​ ∫ 3 15 ​ f(u)du=7∫ 3 15 ​ f(u)du=21.NcW we can write the integral as ∫ 3 15 ​ f(x)dx=21. Here you have to rememben that ∫ 3 5 ​ f(x)dx. Explanation Here actually we made a transformation of variable u=3x, that makes the integral as a different form.

Answer 1

Given the integral ∫₁⁵ f(3x)dx = 7, we make the substitution u = 3x. This transforms the integral into ∫₃¹₅ f(u) * (1/3)du = 7. Solving for the integral in terms of u yields ∫₃¹₅ f(u)du = 21. Therefore, ∫₃¹₅ f(x)dx = 21.

Step-by-step explanation:

Start with the integral ∫₁⁵ f(3x)dx = 7. To simplify, perform the substitution u = 3x. The differential change dx becomes (1/3)du. Substitute these into the integral:

∫₁⁵ f(u) * (1/3)du = 7.

Now, solve for the integral in terms of u:

∫₃¹₅ f(u)du = 21.

However, we need the integral in terms of x, not u. Since u = 3x, substitute back:

∫₃¹₅ f(x)dx = 21.

This is the final result. The key step is recognizing the need for a substitution to simplify the integral. The choice of u = 3x is crucial for making the calculation manageable. This process showcases the flexibility and utility of substitution techniques in integral calculus, allowing us to express the integral in a more convenient form and arrive at the final answer of ∫₃¹₅ f(x)dx = 21.