Solve ∫ (from 2 to 11) of (7x^2+4e^x)1

$$\int _ { 2 } ^ { 11 } ( 7 x ^ { 2 } + 4 e ^ { x } ) 1$$

Evaluate

$$4e^{11}-4e^{2}+3087\approx 242554.010636395$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 7x^{2}+4e^{x}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 7x^{2}\mathrm{d}x+\int 4e^{x}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$7\int x^{2}\mathrm{d}x+4\int e^{x}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 7 times \frac{x^{3}}{3}.

$$\frac{7x^{3}}{3}+4\int e^{x}\mathrm{d}x$$

Use \int e^{x}\mathrm{d}x=e^{x} from the table of common integrals to obtain the result.

$$\frac{7x^{3}}{3}+4e^{x}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{7}{3}\times 11^{3}+4e^{11}-\left(\frac{7}{3}\times 2^{3}+4e^{2}\right)$$

Simplify.

$$3087+4e^{11}-4e^{2}$$