Find the indefinite integral. ∫t^2 ·4^t d t

Name: Problem 17, Chapter 8: Calculus with Concepts in Calculus
Uploaded: 2022-04-06T11:43:23-08:00
Duration: 8 min 42 s
Channel: Sirat Shah
Description: Find the indefinite integral. ∫t^2 ·4^t d t

Key Concepts

Substitution Method

The substitution method is a technique used to simplify an integral by changing the variable of integration. It is especially effective when the integrand contains a composite function, allowing the problem to be reformulated into a more standard form that is easier to integrate.

Indefinite Integration

Indefinite integration is the process of finding an antiderivative or a family of functions whose derivative is the given function. It reverses differentiation and includes an arbitrary constant since antiderivatives are only determined up to an additive constant.

Integration by Parts

Integration by parts is a technique for integrating products of two functions by using the formula ?u dv = uv ? ?v du. It is particularly useful when one function in the product simplifies when differentiated, making the resulting integral easier to evaluate.

Exponential Function Integration

Exponential function integration involves techniques tailored to functions with exponential components. This often includes rewriting exponentials in a form that facilitates the application of integration methods, such as substitution or integration by parts, especially when combined with polynomial or other functions.

Transcript

00:01 In this problem, we have to find integral of t squared times 4 raised to the power t, d of t using integration by parts.

00:19 As we know that integral of udiv is equal to u times v minus integral of v of v times d of u.

00:33 Let us choose u is equal to t square and dv is equal to 4 r to the power t d of because for these choices of u and dv integral v d u is easy to integrate as compared to the original integral now we are going to find v and d of u taking differential of u we get differential of u is equal to differential of t square will be 2 t d of t and to find we are going to integrate dv is equal to integral of 4 raised to the power t d of t now integral of dv is equal to v and integral of 4 raised to the power t will be 4 raise to the power t over natural log of 4 and we use the constant of integration with the final answer.

02:01 Now using these substitutions in the above integration by parts formula we have integral of t square times 4 raised to the power t, d of t is equal to u times v, since u is t squared and v is 4 raised to the power t over natural log of 4 minus integral of v, which is 4 raised to the power t divided by natural log of 4.

02:44 Times d of u, which is 2 times t d of.

02:55 Now, this will be t squared times 4 raised to the power t, divided by natural log of 4, minus taking 2 outside the integral sign, we have 2 over natural log of 4 can be written as 2 times natural log of 2.

03:17 Integral of t times 4 raised to the power t d of t now here 2 2 becomes 1 and we are going to use integration by parts again to find integral of t times 4 raise to the power t d of t for this, consider u is equal to t and dv is equal to 4 race to the power t, d of t.

04:00 This implies that d u is equal to d t and to find v, we have to integrate integral dv is equal to integral of 4 raised to the power t d of t now integral dv is equal to v and integral of 4 r s to the power t is 4 raised to the power t over natural log of 4 now using integration by parts equation 1 can be written as t square 4 race to the power t over natural log of 4 can be written as 2 times natural log of 2 minus 1 over natural log of 2 times u times v since u is t and v is 4 raised to the power t over natural log of 4 minus integral of v which is 4 raised to the power t over natural log of 4 times d u which is equal to d of...

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