Please help me calculate these integrals.

Suppose that Int 0->1 f(t)dt=11 . Calculate each of the following.
A. int0->0.25 f(4t)dt=

B. int0->0.25 f(1−4t)dt=

C. int0.25->0.375 f(3−8t)dt=

To calculate the given integrals, we can use the substitution method. Let's solve each integral step by step:

A.
Given: ∫₀⁰.²⁵ f(4t) dt

We need to substitute u = 4t, which gives us du = 4 dt.
When t = 0, u = 4 × 0 = 0, and when t = 0.25, u = 4 × 0.25 = 1.

The integral becomes:
∫₀¹ f(u) (1/4) du

The definite integral from 0 to 1 with respect to u is still 11.
Therefore, the answer is (1/4)(11) = 11/4.

B.
Given: ∫₀⁰.²⁵ f(1−4t) dt

We need to substitute u = 1 - 4t, which gives us du = -4 dt.
When t = 0, u = 1 - 4(0) = 1, and when t = 0.25, u = 1 - 4(0.25) = 0.

The integral becomes:
∫₁⁰ f(u) (-1/4) du

The definite integral from 1 to 0 with respect to u is equal to the negative of the definite integral from 0 to 1. Hence, it is -11.

Therefore, the answer is (-1/4)(-11) = 11/4.

C.
Given: ∫₀.²⁵⁰.³⁷⁵ f(3−8t) dt

We need to substitute u = 3 - 8t, which gives us du = -8 dt.
When t = 0.25, u = 3 - 8(0.25) = 1, and when t = 0.375, u = 3 - 8(0.375) = 0.

The integral becomes:
∫₁⁰ f(u) (-1/8) du

The definite integral from 1 to 0 with respect to u is again equal to the negative of the definite integral from 0 to 1. Hence, it is -11.

Therefore, the answer is (-1/8)(-11) = 11/8.

To calculate these integrals, we can use the concept of change of variables. By substituting a new variable and applying the change of variable formula, we can rewrite the integrals in a way that will make them easier to evaluate.

Let's start with the first integral: ∫₀^(0.25) f(4t) dt.

A. To evaluate this integral, we need to substitute a new variable. Let u = 4t.

The limits of integration also need to be adjusted accordingly. When t = 0, u = 4(0) = 0, and when t = 0.25, u = 4(0.25) = 1.

Differentiating both sides of the equation u = 4t with respect to t gives du/dt = 4, which implies dt = du/4.

Now, we can rewrite the integral using the new variable u:

∫₀^(0.25) f(4t) dt = ∫₀¹ f(u) * (1/4) du.

Since we know that ∫₀¹ f(u) du = 11 (given), we can substitute this value into the equation:

∫₀^(0.25) f(4t) dt = (1/4) * ∫₀¹ f(u) du = (1/4) * 11 = 11/4.

B. Now, let's move to the second integral: ∫₀^(0.25) f(1-4t) dt.

In this case, let's substitute a new variable v = 1-4t.

When t = 0, v = 1 - 4(0) = 1, and when t = 0.25, v = 1 - 4(0.25) = 0.

Differentiating both sides of the equation v = 1 - 4t with respect to t gives dv/dt = -4, which implies dt = -dv/4.

Now, we can rewrite the integral using the new variable v:

∫₀^(0.25) f(1-4t) dt = ∫₁⁰ f(v) * (-1/4) dv.

Since we know that ∫₀¹ f(v) dv = 11 (given), we can substitute this value into the equation:

∫₀^(0.25) f(1-4t) dt = (-1/4) * ∫₁⁰ f(v) dv = (-1/4) * 11 = -11/4.

C. Finally, let's move to the third integral: ∫₀.²⁵⁸ f(3-8t) dt.

In this case, let's substitute a new variable w = 3-8t.

When t = 0.25, w = 3 - 8(0.25) = 1, and when t = 0.375, w = 3 - 8(0.375) = 0.

Differentiating both sides of the equation w = 3 - 8t with respect to t gives dw/dt = -8, which implies dt = -dw/8.

Now, we can rewrite the integral using the new variable w:

∫₀.²⁵⁸^(0.375) f(3-8t) dt = ∫₃¹⁄₈⁰ f(w) * (-1/8) dw.

To evaluate this integral, we need to find the value of ∫₁⁰ (3-8t) dt. Let's substitute w = 3 - 8t into this integral:

∫₁⁰ (3-8t) dt = ∫₁⁰ (3-8t) * (-1/8) dw.

The limits of integration stay the same because w = 3 - 8t:

∫₁⁰ (3-8t) dt = ∫₁⁰ (3-8t) * (-1/8) dw = ∫₃¹⁄₈⁰ (3/8 - w/8) dw.

Splitting this integral into two parts gives:

∫₃¹⁄₈⁰ (3/8 - w/8) dw = (3/8) ∫₃¹⁄₈⁰ dw - (1/8) ∫₃¹⁄₈⁰ w dw.

Evaluating each integral:

(3/8) ∫₃¹⁄₈⁰ dw = (3/8) * (0.375-0.258) = (3/8) * 0.117 = 0.043875.

(1/8) ∫₃¹⁄₈⁰ w dw = (1/8) * [(0.375)² - (0.258)²] = (1/8) * (0.140625 - 0.066564) = 0.010805.

Now, we can substitute these results back into the original equation:

∫₀.²⁵⁸^(0.375) f(3-8t) dt = ∫₃¹⁄₈⁰ f(w) * (-1/8) dw = -(3/8) + (1/8) * 0.010805 = -0.328125 + 0.001351 = -0.326774.