Solve ∫ (from 1 to 4) of x^2(x^4-7)^3

$$\int _ { 1 } ^ { 4 } x ^ { 2 } ( x ^ { 4 } - 7 ) ^ { 3 }$$

Evaluate

$$\frac{3515177736}{55}\approx 63912322.472727273$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x^{4}-7\right)^{3}.

$$\int _{1}^{4}x^{2}\left(\left(x^{4}\right)^{3}-21\left(x^{4}\right)^{2}+147x^{4}-343\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 4 and 3 to get 12.

$$\int _{1}^{4}x^{2}\left(x^{12}-21\left(x^{4}\right)^{2}+147x^{4}-343\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$\int _{1}^{4}x^{2}\left(x^{12}-21x^{8}+147x^{4}-343\right)\mathrm{d}x$$

Use the distributive property to multiply x^{2} by x^{12}-21x^{8}+147x^{4}-343.

$$\int _{1}^{4}x^{14}-21x^{10}+147x^{6}-343x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{14}-21x^{10}+147x^{6}-343x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{14}\mathrm{d}x+\int -21x^{10}\mathrm{d}x+\int 147x^{6}\mathrm{d}x+\int -343x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{14}\mathrm{d}x-21\int x^{10}\mathrm{d}x+147\int x^{6}\mathrm{d}x-343\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{14}\mathrm{d}x with \frac{x^{15}}{15}.

$$\frac{x^{15}}{15}-21\int x^{10}\mathrm{d}x+147\int x^{6}\mathrm{d}x-343\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{10}\mathrm{d}x with \frac{x^{11}}{11}. Multiply -21 times \frac{x^{11}}{11}.

$$\frac{x^{15}}{15}-\frac{21x^{11}}{11}+147\int x^{6}\mathrm{d}x-343\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply 147 times \frac{x^{7}}{7}.

$$\frac{x^{15}}{15}-\frac{21x^{11}}{11}+21x^{7}-343\int x^{2}\mathrm{d}x$$

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 -343 times \frac{x^{3}}{3}.

$$\frac{x^{15}}{15}-\frac{21x^{11}}{11}+21x^{7}-\frac{343x^{3}}{3}$$

Simplify.

$$-\frac{343x^{3}}{3}+21x^{7}-\frac{21x^{11}}{11}+\frac{x^{15}}{15}$$

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.

$$-\frac{343}{3}\times 4^{3}+21\times 4^{7}-\frac{21}{11}\times 4^{11}+\frac{4^{15}}{15}-\left(-\frac{343}{3}\times 1^{3}+21\times 1^{7}-\frac{21}{11}\times 1^{11}+\frac{1^{15}}{15}\right)$$

Simplify.

$$\frac{3515177736}{55}$$