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Chapter 5: Problem 101

Simplify the given expressions. $$\frac{d}{d x} \int_{0}^{\cos x}\left(t^{4}+6\right) d t$$

Short Answer

Expert verified
Question: Simplify the expression given by the derivative of the integral with respect to x. Expression: $$\frac{d}{d x} \int_{0}^{\cos x} (t^4 + 6) dt$$ Answer: $$(-\sin x)((\cos x)^4 + 6)$$

Step by step solution

01

Identify the function and its derivative

In this case, the function we're integrating is \(f(t) = t^4 + 6\). The variable we're differentiating with respect to is \(x\). The derivative of \(f(t)\) with respect to \(t\) is: $$f'(t) = \frac{d}{dt}(t^4 + 6) = 4t^3$$
02

Differentiate the bounds of the integral

Since the lower bound is 0, which is a constant, its derivative is 0. The upper bound is \(\cos x\). To apply the fundamental theorem of calculus, we need to differentiate the upper bound with respect to \(x\). $$\frac{d}{dx} (\cos x) = -\sin x$$
03

Apply the Fundamental Theorem of Calculus

Now we can apply the fundamental theorem of calculus to simplify the expression: $$\frac{d}{d x} \int_{0}^{\cos x}\left(t^{4}+6\right) d t = f(\cos x)\cdot (-\sin x) = \left((\cos x)^4 + 6\right)\cdot (-\sin x)$$ So, the simplified expression is: $$\boxed{(-\sin x)((\cos x)^4 + 6)}.$$

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Most popular questions from this chapter

Average value of sine functions Use a graphing utility to verify that the functions \(f(x)=\sin k x\) have a period of \(2 \pi / k,\) where \(k=1,2,3, \ldots . .\) Equivalently, the first "hump" of \(f(x)=\sin k x\) occurs on the interval \([0, \pi / k] .\) Verify that the average value of the first hump of \(f(x)=\sin k x\) is independent of \(k .\) What is the average value?

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Evaluate the following definite integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{4} \frac{x-2}{\sqrt{x}} d x$$
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