Q. 68 Prove Theorem 4.35 in your own w... [FREE SOLUTION]

Chapter 4: Q. 68 (page 401)

Prove Theorem 4.35 in your own words: If $f$ is continuous on $[a, b]$ and $u (x)$ is a differentiable function, then for all $x 鈭� [a, b]$ , role="math" localid="1648747196651" ${鈭�}_{a}^{u (x)} f (t) d t = f (u (x)) u' (x)$ . Be especially clear about how you use the chain rule.

Short Answer

Expert verified

If $f$ is continuous on $[a, b]$ and $u (x)$ is a differentiable function, then for all $x 鈭� [a, b]$ , role="math" localid="1648747371503" ${鈭�}_{a}^{u (x)} f (t) d t = f (u (x)) u' (x)$ .

Step by step solution

Step 1. Given information

We have to prove ${鈭�}_{a}^{u (x)} f (t) d t = f (u (x)) u' (x)$ .

Step 2. Proof of the given question.

Let $F$ be an antiderivative of $f$ .

Now,

$\frac{d}{d x} {鈭�}_{a}^{u (x)} f (t) d t = \frac{d}{d x} (F (u (x))) = \frac{d}{d x} (F (u (x))) 脳 \frac{d}{d x} (u (x)) = F' (u (x)) u' (x) = f (u (x)) u' (x)$

Therefore, it is proved.

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91影视

Prove Theorem 4.35 in your own words: If $f$ is continuous on $[a, b]$ and $u (x)$ is a differentiable function, then for all $x 鈭� [a, b]$ , role="math" localid="1648747196651" ${鈭�}_{a}^{u (x)} f (t) d t = f (u (x)) u' (x)$ . Be especially clear about how you use the chain rule.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proof of the given question.

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91影视

Prove Theorem 4.35 in your own words: If fis continuous on [a,b]and uxis a differentiable function, then for all x鈭�a,b, role="math" localid="1648747196651" 鈭�auxftdt=fuxu'x. Be especially clear about how you use the chain rule.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proof of the given question.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Pure Maths

Probability and Statistics

Applied Mathematics

Decision Maths

Geometry

Study anywhere. Anytime. Across all devices.

Prove Theorem 4.35 in your own words: If $f$ is continuous on $[a, b]$ and $u (x)$ is a differentiable function, then for all $x 鈭� [a, b]$ , role="math" localid="1648747196651" ${鈭�}_{a}^{u (x)} f (t) d t = f (u (x)) u' (x)$ . Be especially clear about how you use the chain rule.