Derivative integral Please Help

Question

Answer 1

Find:

Solution:

Let's define some function:

G(x) = ∫₀^x exp( cos( 4t ) ) dt

So if I wanted to do:

G( 12x + 2000 ), then G( 12x + 2000 ) = ∫₀^{12x + 2000} exp( cos( 4t ) ) dt

Weird example, I know, but I hope you catch my point that we can put ANYTHING in the upper bound for the integral now

And also:

G'( u(x) ) = exp( cos( 4u(x) ) ) * u'(x)

By Chain Rule and the Fundamental Theorem of Calculus

So, some other F(x) would be defined as:

F(x) = ∫_-15x^2^x-3 exp( cos( 4t ) ) dt

Let's rewrite F(x) so that we can take advantage of G(x)

F(x) = ∫₀^x-3 exp( cos( 4t ) ) dt + ∫_-15x^2⁰ exp( cos( 4t ) ) dt

F(x) = ∫₀^x-3 exp( cos( 4t ) ) dt - ∫₀^{-15x^2} exp( cos( 4t ) ) dt

Thus,

F(x) = G( x - 3 ) - G( -15x² )

And,

d/dx( F(x) ) = G'( x - 3 ) * +1 - G'( -15x² ) * -30x

d/dx( F(x) ) = exp( cos( 4( x - 3 ) ) + 30x * exp( cos( 4( -15x² ) )

∴ d/dx( F(x) ) = exp( cos( 4x - 12 ) ) + 30x * exp( cos( 60x² ) )

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!

Answer 2

Integrate from -15x² to 0 and then add the integral from 0 to -x+3

Then apply the Fundamental Theorem of Calculus.: d/dx[∫_{(from a to u)}g(t)dt] = g(u) u' (a is a constant)