Math Problem Statement

(x^2+4)e^-xdx from 0 to 2

Solution

We need to evaluate the integral of the function (x2+4)ex(x^2 + 4)e^{-x} from 0 to 2. This is a definite integral, and we can solve it using integration by parts. Let's break it down step by step.

The integral to evaluate is:

I=02(x2+4)exdxI = \int_0^2 (x^2 + 4) e^{-x} \, dx

Step 1: Split the integral

We can break the integral into two parts:

I=02x2exdx+024exdxI = \int_0^2 x^2 e^{-x} \, dx + \int_0^2 4 e^{-x} \, dx

Let's solve these two integrals separately.

Step 2: Solving the first integral 02x2exdx\int_0^2 x^2 e^{-x} \, dx

We use integration by parts. Let’s first choose the parts:

  • Let u=x2u = x^2, which means du=2xdxdu = 2x \, dx.
  • Let dv=exdxdv = e^{-x} \, dx, so v=exv = -e^{-x}.

Using the integration by parts formula:

udv=uvvdu\int u \, dv = uv - \int v \, du

Now, apply this to x2exdx\int x^2 e^{-x} \, dx:

x2exdx=x2ex02+2xexdx\int x^2 e^{-x} \, dx = -x^2 e^{-x} \Big|_0^2 + 2 \int x e^{-x} \, dx

For xexdx\int x e^{-x} \, dx, apply integration by parts again:

  • Let u=xu = x, so du=dxdu = dx.
  • Let dv=exdxdv = e^{-x} \, dx, so v=exv = -e^{-x}.

xexdx=xex02+exdx\int x e^{-x} \, dx = -x e^{-x} \Big|_0^2 + \int e^{-x} \, dx xexdx=xex02+ex02\int x e^{-x} \, dx = -x e^{-x} \Big|_0^2 + e^{-x} \Big|_0^2

Evaluating both integrals at the limits:

xex02=(2e20e0)=2e2-x e^{-x} \Big|_0^2 = -(2e^{-2} - 0e^{-0}) = -2e^{-2} ex02=e2e0=e21e^{-x} \Big|_0^2 = e^{-2} - e^{0} = e^{-2} - 1

Thus:

xexdx=2e2+(e21)=e21\int x e^{-x} \, dx = -2e^{-2} + (e^{-2} - 1) = -e^{-2} - 1

Now, substitute back into the expression for x2exdx\int x^2 e^{-x} \, dx:

x2exdx=x2ex02+2(e21)\int x^2 e^{-x} \, dx = -x^2 e^{-x} \Big|_0^2 + 2(-e^{-2} - 1) x2ex02=(4e20e0)=4e2-x^2 e^{-x} \Big|_0^2 = -(4e^{-2} - 0e^{-0}) = -4e^{-2}

Thus:

02x2exdx=4e2+2(e21)=4e22e22=6e22\int_0^2 x^2 e^{-x} \, dx = -4e^{-2} + 2(-e^{-2} - 1) = -4e^{-2} - 2e^{-2} - 2 = -6e^{-2} - 2

Step 3: Solving the second integral 024exdx\int_0^2 4 e^{-x} \, dx

This is a straightforward integral:

4exdx=4ex02=4(e2e0)=4(e21)\int 4 e^{-x} \, dx = -4 e^{-x} \Big|_0^2 = -4(e^{-2} - e^{0}) = -4(e^{-2} - 1) =4(1e2)= 4(1 - e^{-2})

Step 4: Combine the results

Now, add the two integrals together:

I=(6e22)+4(1e2)I = (-6e^{-2} - 2) + 4(1 - e^{-2}) I=6e22+44e2I = -6e^{-2} - 2 + 4 - 4e^{-2} I=210e2I = 2 - 10e^{-2}

Thus, the value of the definite integral is:

210e2\boxed{2 - 10e^{-2}}

Would you like more details on any of the steps, or have any questions?

Follow-up questions:

  1. What is the integration by parts method, and how is it derived?
  2. How do definite integrals differ from indefinite integrals?
  3. What are common strategies for dealing with exponential functions in integrals?
  4. Can you explain why we chose u=x2u = x^2 in the first integration by parts step?
  5. How would this integral change if we replaced exe^{-x} with exe^{x}?

Tip:

Always double-check your integration limits when calculating definite integrals to avoid sign errors in your final result.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integrals
Integration by Parts
Exponential Decay Functions

Formulas

Integration by parts formula: ∫u dv = uv - ∫v du
Exponential integral: ∫e^(-x) dx = -e^(-x)

Theorems

Fundamental Theorem of Calculus
Integration by Parts Theorem

Suitable Grade Level

Grades 11-12 or Early College