How do you use the first part of the Fundamental Theorem of Calculus to find the derivative of $y = \int 3 {(sin (t))}^{4} d t$ $y = int 3(sin(t))^4 dt$ from $e^{x}$ $e^x$ to 1?

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Answer 1 · 2018-07-23T12:46:02

The derivative is $- 3 e^{x} {(sin (e^{x}))}^{4}$ $-3e^x(sin(e^x))^4$ .

$y = \int_{e^{x}}^{1} 3 {(sin (t))}^{4} d t$ $y = int_(e^x)^(1)3(sin(t))^4dt$

Let's let $f (t) = 3 {(sin (t))}^{4}$ $f(t) = 3(sin(t))^4$ .

Remember that by the fundamental theorem of calculus, a definite integral of a function is calculated using its antiderivative.

$\int_{a}^{b} f (t) d t = F (b) - F (a)$ $int_a^b f(t) dt = F(b) - F(a)$ where $F$ $F$ is an antiderivative of $f$ $f$ .

Similarly:

$\int_{e^{x}}^{1} f (t) d t = F (1) - F (e^{x})$ $int_(e^x)^1 f(t)dt = F(1) - F(e^x)$ where $F$ $F$ is an antiderivative of $f$ $f$ .

$y = F (1) - F (e^{x})$ $y = F(1) - F(e^x)$

Because the question asks for it, we take the derivative.

$\frac{d}{d x} y = \frac{d}{d x} (F (1)) - \frac{d}{d x} (F (e^{x}))$ $color(red)(d/dx)y = color(red)(d/dx)(F(1)) - color(red)(d/dx)(F(e^x))$

Don't be scared. $F (1)$ $F(1)$ is just a constant.

$dy/dx = 0 - F'(e^x)*d/dx(e^x)$

But by definition of an antiderivative, $F'(x) = f(x)$ .

$dy/dx = -e^xf(e^x)$

$dy/dx = -e^x(3(sin(e^x))^4)$

$dy/dx = -3e^x(sin(e^x))^4$

How do you use the first part of the Fundamental Theorem of Calculus to find the derivative of y = int 3(sin(t))^4 dty=∫3(sin(t))4dty = int 3(sin(t))^4 dt from e^xexe^x to 1?